LIBRARY 

OF  THE 

UNIVERSITY  OF  CALIFORNIA. 


Class 


THE  ELECTRIC  CIRCUIT 


BY 

V.  KARAPETOFF 


-»      •>      *>    ^      ** 


ITHACA,   NEW   YORK 
1910 


Engineering 
library 


COPYRIGHT.  1910 
BY   V.  KARAPEfOFF 


PRESS   OF 

ANI>RUS  &  CHURCH 
ITHACA,  N.  Y. 


PREFACE. 


This  pamphlet,  together  with  the  companion  pamphlet  entitled 
"The  Magnetic  Circuit,"  is  intended  to  give  a  student  in 
electrical  engineering  the  theoretical  elements  necessary  for 
calculation  of  the  performance  of  dynamo-electric  machinery  and 
of  transmission  lines.  The  advanced  student  must  be  taught  to 
treat  every  electric  machine  as  a  particular  combination  of  electric 
and  magnetic  circuits,  and  to  base  its  performance  upon  the 
fundamental  theoretical  relations  rather  than  upon  a  separate 
"theory"  established  for  each  kind  of  machinery,  as  is  often 
done. 

The  first  chapter  is  devoted  to  a  review  of  the  direct- current 
circuit,  the  next  four  chapters  treat  of  sine-wave  alternating- 
current  circuits,  and  the  last  two  chapters  give  the  fundamental 
properties  of  the  electrostatic  circuit.  All  the  important  results 
and  methods  are  illustrated  by  numerical  problems  of  which  there 
are  over  one  hundred  in  the  text.  The  pamphlet  is  not  intended 
for  a  beginner,  but  for  a  student  who  has  had  an  elementary 
descriptive  course  in  electrical  engineering  and  some  simple 
laboratory  experiments. 

The  treatment  is  made  as  far  as  possible  uniform,  so  that  the 
student  sees  analogous  relations  in  the  direct-current  circuit,  in 
the  alternating-current  circuit,  in  the  electrostatic  circuit,  and 
finally  in  the  magnetic  circuit.  All  matter  of  purely  historical 
or  academic  interest,  not  bearing  directly  upon  the  theory  of 
electric  machinery,  has  been  left  out.  An  ambitious  student  will 
find  a  more  exhaustive  treatment  in  the  works  mentioned  at  the 
end  of  the  pamphlet. 

The  electrostatic  circuit  is  treated  in  accordance  with  the 
modern  conception  of  elastic  displacement  of  electricity  in 
dielectrics.  No  use  has  been  made  of  the  action  of  electric 
charges  at  a  distance,  or  of  the  electrostatic  system  of  units. 
The  volt-ampere-ohm  system  of  units  is  used  for  electrostatic 
calculations,  in  accordance  with  Professor  Giorgi's  ideas  (see  a 
paper  by  Professor  Ascoli  in  Vol.  I  of  the  Transactions  of  the 


225063 


iv  PREFACE. 

International  Electrical  Congress,  St.  Louis,  1904).  Those 
familiar  with  Oliver  Heaviside's  writings  will  notice  his  influence 
upon  the  author,  in  particular  in  Arts.  22  and  23,  where  an 
attempt  is  made  at  a  rational  electrostatic  nomenclature. 

Many  thanks  are  due  to  the  author's  friend  and  colleague,  Mr. 
John  F.  H.  Douglas,  instructor  in  electrical  engineering  in 
Sibley  College,  who  read  the  manuscript  and  the  proofs,  checked 
the  answers  to  the  problems,  and  made  many  excellent  suggestions 
for  the  text. 

Cornell  University,  Ithaca,  N.  Y. 
August 


TABLE    OF   CONTENTS. 


CHAPTER  I.    ELECTRICAL,  RELATIONS  IN  DIRECT-CURRENT  CIRCUITS. 

PAGE 

1.  The  Volt,  the  Ampere,  and  the  Ohm. i 

2.  Resistances  and  Conductances  in  Series  and  in  Parallel 3 

3.  Resistivity  and  Conductivity 6 

4.  Current  Density  and  Electric  Intensity 8 

5.  Conductors  of  Variable  Cross-section 9 

CHAPTER  II.    REPRESENTATION  OF  ALTERNATING  CURRENTS  AND 
VOLTAGES  BY  SINE  WAVES  AND  BY  VECTORS. 

6.  Sinusoidal  Voltages  and  Currents 13 

7.  Representation  of  a  Sine-wave  by  a  Vector 17 

8.  Addition  and  Subtraction  of  Vectors 20 

CHAPTER  III.    POWER  IN  ALTERNATING-CURRENT  CIRCUITS. 

9.  Power  when  Current  and  Voltage  are  in  Phase. — Effective  value 

of  sine  wave 25 

10.  Power  when  Current  and  Voltage  are  out  of  Phase. — Wattless  and 

Energy  Components. — Form  factor 30 

CHAPTER  IV.    REACTANCE  AND  RESISTANCE  IN  ALTERNATING- 
CURRENT  CIRCUITS. 

11.  Inductance  and  Reactance 35 

12.  Problems  on  Impedance,  and  Impedances  in  Series 37 

13.  Impedances  in  Parallel 38 

14.  Equivalent  Resistance  and  Reactance 39 

15.  Admittance  and  its  Components — Conductance  and  Susceptance 41 

CHAPTER  V.    THE  USE  OF  COMPLEX  QUANTITIES. 

16.  Addition  and  Subtraction  of  Vectors  in  Projections 45 

17.  Rotation  of  Vectors  by  Ninety  Degrees 47 

18.  Impedance  and  Admittance   Expressed  as   Complex  Quantities. — 

Abbreviated  notation 50 

19.  Power  and  Phase  Displacement  Expressed  Through  Projections  of 

Vectors 55 

CHAPTER  VI.    THE  ELECTROSTATIC  CIRCUIT. 

20.  Physical  Concept  of  the  Electrostatic  Field 58 

21.  Dielectric     Flux     Density     and     Electrostatic     Stress     (Voltage 

Gradient) 63 

22.  Permittivity  and  Dielectric  Strength 67 

23.  Permittance,  or  Electrostatic  Capacity 69 

CHAPTER  VII.    THE  ELECTROSTATIC  CIRCUIT.  (Continued.) 

24.  Capacity  in  Alternating-Current  Circuits 76 

25.  Energy  in  the  Electrostatic  Field 78 

26.  The  Electf  ostatic  Corona 82 

27.  Dielectric  Hysteresis  and  Conduction 84 

LITERATURE. 


CHAPTER  I. 

ELECTRICAL  RELATIONS  IN  DIRECT-CURRENT 
CIRCUITS. 

i.  The  Volt,  the  Ampere,  and  the  Ohm.  The  student  is 
supposed  to  be  familiar  with  Ohm's  law,  both  theoretically  and 
from  his  laboratory  experience.  A  brief  synopsis  of  the  law, 
given  below,  is  intended  merely  to  refresh  the  relations  in  his 
mind,  and  to  establish  a  point  of  view  which  permits  of  extend- 
ing these  relations  to  alternating-current  circuits.  Moreover, 
the  law  is  presented  in  a  form  applicable  to  the  magnetic  and  to 
the  dielectric  circuits. 

When  the  current  in  a  conductor  is  steady,  its  value  is  propor- 
tional to  the  voltage  at  the  terminals  of  the  conductor.  This  is 
an  experimental  fact,  called  Ohm's  law.  Considering  the  electro- 
motive force  as  the  cause  of  the  current,  this  law  merely  states 
that  the  effect  is  proportional  to  the  cause,  or 

*-r-i (i) 

where  the  coefficient  of  proportionality  r  is  called  the  resistance 
of  the  conductor.  When  the  current  is  expressed  in  amperes, 
and  the  electro-motive  force  in  volts,  the  resistance  r  is  measured 
in  units  called  ohms. 

With  our  present  meager  knowledge  of  the  true  nature  of 
electrical  phenomena  it  is  well-nigh  impossible  to  give  a  clear 
physical  meaning  of  the  quantities  under  discussion,  without 
resorting  to  analogies.  For  instance,  the  flow  of  current  through 
a  conductor  may  be  compared  to  the  flow  of  heat  through  a  rod, 
the  voltage  being  analogous  to  the  difference  of  temperatures  at 
the  ends  of  the  rod,  and  the  electric  current  to  the  quantity  of 
heat  passing  through  a  cross-section  of  the  rod  in  unit  time  (rate 
of  flow  of  heat).  The  ratio  of  the  two  can  be  called  the  resistance 
of  the  rod  to  the  passage  of  heat.  Again,  the  phenomenon  of 
flow  of  electricity  is  somewhat  analogous  to  the  flow  of  water 
through  pipes.  The  hydraulic  head  may  be  likened  to  the  voltage, 
and  the  rate  of  discharge  of  water  to  the  current.  With  very 
low  velocities,  in  capillary  tubes,  the  discharge  is  proportional 
to  the  head,  so  that  eq.  (i)  holds  true  for  the  flow  of  water. 


2  DIRECT-CURRENT  CIRCUITS.  [CHAP,  i 

Whatever  the  reasons  which  have  originally  led  to  the  choice 
of  the  magnitudes  of  the  ampere,  the  ohm,  and  the  volt,  these 
units  may  be  considered  at  present,  for  all  practical  and  most 
theoretical  purposes,  as  arbitrary  units,  same  as  the  foot,  the 
pound,  or  the  meter.  Their  values  have  been  established  by 
international  agreements,  whence  the  name  :  international  elec- 
trical units.  These  units  are  represented  by  concrete  standards 
with  minutely  specified  dimensions  and  properties  :  The  ohm  by  a 
column  of  mercury,  the  ampere  by  a  silver  voltameter,  and  the 
volt  by  standard  cells.  It  is  understood,  of  course,  that  only 
two  out  of  the  three  units  need  to  be  standardized,  the  third 
being  determined  either  as  their  product,  or  their  ratio.  At  this 
writing,  there  is  some  discussion,  as  to  whether  the  ampere  or 
the  volt  should  be  legalized.  This  discussion  is  of  no  immediate 
importance  to  the  engineer,  since  both  the  volt  and  the  ampere 
are  known  with  an  accuracy  far  exceeding  his  needs. 

The  ampere,  the  volt,  and  the  ohm  are  connected  by  simple 
multipliers  (powers  of  10)  with  the  absolute  electro-magnetic 
units,  or  the  so-called  C.G.S.  system  of  units.  It  is  conceded 
at  present  by  some  prominent  physicists  that  the  choice  of  the 
units  was  not  quite  fortunate,  according  to  our  present  under- 
standing of  the  electro-magnetic  relations.  Since,  however,  it  is 
too  late  to  change  these  units,  it  is  better  to  consider  them  as 
arbitrary  units,  not  connected  in  any  way  with  the  magnitudes 
of  the  centimeter,  the  gramme,  and  the  second. 

In  applying  Ohm's  law  to  practical  problems  it  must  be  re- 
membered that  e  represents  the  net  voltage  at  the  ends  of  the 
conductor  r.  It  is  necessary  to  keep  this  in  mind  when  the  cir- 
cuit contains  sources  of  counter-electromotive  forces,  such  as 
electric  batteries,  or  motors.  Let,  for  instance,  the  total  resist- 
ance of  a  circuit,  connected  across  the  terminals  of  a  generator, 
be  12  ohms,  and  let  the  terminal  voltage  of  the  generator  be  120 
volts.  Then  the  current  is  equal  to  10  amperes,  provided  that 
there  are  no  counter-electromotive  forces  in  the  circuit.  Let, 
however,  the  circuit  contain  a  storage  battery  of  say  24  volts, 
connected  so  as  be  charging,  that  is  opposing  the  applied  voltage. 
The  current  in  the  circuit  is  now  only  (120  —  24)!  12  =  8  amps., 
the  value  120 — 24  =  96  being  the  net  voltage  in  the  external 
circuit.  Should  the  terminals  of  the  battery  be  reversed,  so  as 


CHAP,  i]  DIRECT-CURRENT  CIRCUITS.  3 

to  help  the  generator  voltage,   the  current  would  increase  to 
(120  -f-  24)/i2  =  12  amps. 

2.  Resistances  and  Conductances  in  Series  and  in 
Parallel.  When  resistances  are  added  in  series,  the  total  resist- 
ance of  the  circuit  increases.  The  reason  is  the  same  for  which 
an  increase  in  the  length  of  a  pipe  makes  its  frictional  resistance 
larger,  or  increasing  the  length  of  a  rod  makes  the  passage  of 
heat  through  it  more  difficult.  The  equivalent  resistance  of  two 
resistances  connected  in  series  is  equal  to  their  sum.  This  fol- 
lows directly  from  the  experimental  fact  that  the  current  is  the 
same  in  all  parts  of  the  same  circuit.  Namely,  let  two  conductors 
rl  and  r2  be  connected  in  series  across  a  source  of  voltage  <?,  and 
let  a  current  i  flow  through  them.  Part  of  the  total  voltage  e  is 
spent  in  overcoming  the  resistance  of  the  first  conductor,  the 
rest  in  overcoming  that  of  the  second  conductor.  But,  according 
to  Ohm's  law,  when  the  conditions  are  steady,  the  voltage  across 
the  first  conductor,  el  —  i  •  rv ;  the  voltage  across  the  second  is 
<?2  =  i  •  r,.  Adding  these  two  equations,  gives  the  total  voltage 
e  =  el  +  e^  =  i(rl  +  rt). 

An  equivalent  conductor,  r^,  by  definition,  is  a  conductor 
which,  with  the  same  total  voltage  <?,  allows  the  same  current  i 
to  pass  through  the  circuit,  as  the  combination  of  the  two 
given  conductors.  Or 

'— *v*V 

Comparing  the  two  foregoing  equations,  gives 

•W-n+.r,  '.         .     .     ,    .    .    (2) 

The  law  can  be  extended  to  any  number  of  conductors  in  series, 
by  gradually  combining  them  into  groups  of  two. 

When  several  conductors  are  connected  in  parallel,  the  voltage 
across  them  is  the  quantity  common  for  all  the  branches,  so  that 
we  have 


(3) 


where  *'„  z'2, are  the  currents  in  the  separate  branches. 

The  total  current  flowing  through  the  given  system  of  conductors 
is  equal  to  the  sum  of  the  currents  in  the  separate  branches. 


4  DIRECT-CURRENT  CIRCUITS.  [CHAP,  i 

This  follows  from  the  experimental  fact  that  electricity  in  its 
flow  behaves  like  an  incompressible  fluid,  that  is  to  say,  the  same 
quantity  of  it  must  pass,  at  each  moment,  through  all  the  cross- 
sections  of  a  circuit.  This  is  again  analogous  to  the  flow  of 
heat,  or  of  water.  Thus,  the  equivalent  conductor,  r,q,  is  deter- 
mined by  the  condition 

'  =  (*!  +  *,+ )•  v  .....  (4) 

Substituting  the  values  of  ilt  z'2,  etc.,  from  (3)  into  (4)  and 
cancelling  e,  gives 

i=L+  L+etc.      .    '.      ...      ,      (5) 

req          r\          r* 

or,  in  words:  when  two  or  more  conductors  are  connected  in 
parallel,  the  reciprocal  of  the  resistance  of  the  equivalent  con- 
ductor is  equal  to  the  sum  of  the  reciprocals  of  the  resistances 
of  the  individual  conductors. 

When  the  resistance  of  one  of  the  conductors  in  parallel  is 
equal  to  zero,  the  corresponding  term  in  eq.  (5)  becomes  infinitely 
large.  The  equivalent  resistance  of  the  whole  combination  is  in 
this  case  equal  to  zero.  This  is  natural  to  expect  because  the 
conductor  of  zero  resistance  short-circuits  the  other  resistances. 

Whenever  resistances  in  parallel  are  to  be  investigated,  their 
reciprocals  enter  into  formulae.  It  is  convenient,  therefore,  to 
define  the  reciprocal  of  a  resistance  as  a  separate  physical 
quantity.  This  is  legitimate  because,  by  definition,  resistance  is 
merely  a  coefficient  of  proportionality  between  the  voltage  and 
the  current.  Ohm's  law  can  be  written  in  the  form 

i  =  g  -e     .     .     .     ,     .      .      .      (6) 
where  the  new  coefficient  of  proportionality 

£"=  i/^ 

is  called  the  conductance  of  the  conductor.  The  reason  for  this 
name  is  easy  to  see :  the  resistance  shows  how  difficult  it  is  to 
force  a  unit  current  through  a  given  conductor,  while  its  recip- 
rocal g  shows  how  easy  it  is  to  produce  the  same  current  in  the 
same  conductor.  Conductances  are  measured  in  units  called 
mhos,  one  mho  being  the  reciprocal  of  one  ohm.  Hence,  a 
standard  resistance,  equal  to  one  ohm,  represents  at  the  same  time 
a  standard  conductance  of  one  mho.  A  resistance  of  two  ohms 


CHAP,  i]  DIRECT-CURRENT  CIRCUITS.  5 

has  a  conductance  of  one  half  mho,  etc.  Increasing  the  resistance 
of  a  winding  from  4  to  5  ohms  reduces  its  conductance  from  .25 
to  .20  mho. 

Equation  (5)  becomes  simply 

^=£"1 +<£•*  + etc.     .     .     .     .   ^    .    (7) 

or  the  equivalent  conductance  of  several  conductors  in  parallel 
is  equal  to  the  sum  of  their  conductances.  As  a  general  rule,  in 
the  solution  of  problems  it  is  convenient  to  use  conductances 
when  conductors  are  in  parallel,  and  to  use  resistances,  when 
they  are  in  series.  In  this  way,  the  necessity  for  introducing 
reciprocal  expressions  into  formulae  is  obviated. 

Power  converted  into  heat  in  a  conductor  is  experimentally 
found  to  be  proportional  to  ei.  In  this  respect  electricity  behaves 
like  a  liquid  in  a  pipe.  There  also  the  power  is  proportional  to 
the  rate  of  flow  times  the  difference  of  pressures  at  the  ends  of 
the  pipe.  The  three  units,  the  volt,  the  ampere,  and  the  watt 
have  been  originally  so  established,  that  the  coefficient  of  pro- 
portionality is  equal  to  unity,  and  watts  =  volts  x  amperes.  If 
other  units  are  used  for  power,  for  instance  kilowatts,  or  horse- 
power, a  coefficient  of  proportionality  is  necessary. 

If  r  is  known,  instead  of  £,  then  substituting  its  value  from 
eq.  (i)  we  find  the  familiar  expression  i*r  for  the  power.  If  it 
is  desired  to  eliminate  t,  the  expression  for  power  becomes  e*\r. 
Using  conductance,  instead  of  resistance,  in  the  last  two  ex- 
pressions, gives  two  additional  formulae  for  power : 

Power  =  i*\g  =  e*g.     .....     (ya) 

PROB.  i.  Two  resistances,  r^  =  5  ohm,  and  r.z  =  7  ohm,  are 
connected  in  series.  Resistance  r^  is  shunted  by  a  comparatively 
high  resistance,  Rl  —  100  ohm. ;  rz  is  shunted  by  a  resistance 
7?a  =  50  ohm.  What  is  the  equivalent  resistance  of  the  whole 
combination  ?  Solution : 

equivalent  conductance  of  rx  and  Rv  is  .2  +  .01  =  .21  mho; 

resistance       "   "     "     ""         i/.2i  =  4.76  ohm ; 
4 '  conductance  of  r2  and  R^  is 

.  1429  4-  .0200  =  .  1629  mho  ; 

resistance       of  r2  and  R2  is    i/.  1629  =  6.14  ohm. 
Ans.  4.76  4-  6.14  =  10.90  ohm. 


6  DIRECT-CURRENT  CIRCUITS.  [CHAP,  i 

PROB.  2.  If  the  currents  in  the  shunted  resistances  Rl  and  Rt 
in  the  preceding  problem  represent  pure  loss  of  power,  what  is 
the  efficiency  of  the  whole  arrangement?  Solution  :  Let  the 
voltage  across  the  resistances  rv  and  Rl  be  e.  Then  the  voltage 
across  rz  and  R^  is  e^  X  6.14/4.76  =  1.29  e.  Hence,  the  useful 
power  is  ^2/5  +  (I-29^)2/7  =  -438  e*  watts.  The  power  lost  in  the 
resistances,  R^  and  R^  is  e*lioo  -f  (i.  29^)^50=  .0433^  watts. 
Therefore,  the  efficiency  is  .438/(.438  +  .0433)  =  90  per  cent. 

PROB.  3.  Four  resistances,  rl  =  1.2,  r2  =  1.7,  r=25,  andro  = 
750  ohms,  are  connected  as  shown  in  Fig.  i  .  The  generator  voltage 
between  the  points  A  and  B  is  500  v.  Determine  the  current 
through  the  resistance  r  and  the  voltage  across  this  resistance. 
NOTE.  This  combination  represents  a  transmission  line  of  a 
resistance  r^  +  rv  the  useful  load  resistance  r,  and  the  leakage 


VvAAAAAA 


B 


. 

FIG.  i.     A  series-parallel  combination  of  resistances. 

resistance  ro.  The  problem  is  of  a  great  importance  in  alternat- 
ing-current circuits  ;  see  problem  96,  Chapter  V.  Solution  : 
Combine  the  resistances  r2  and  r  into  one  ;  determine  the  con- 
ductance il(r  +  r,),  and  combine  it  with  the  leakage  conductance 
i/r0.  Determine  the  equivalent  resistance  between  the  points  C 
and  D,  and  the  total  resistance  between  A  and  B.  Having  found 
the  total  current,  subtract  from  the  generator  voltage  the 
voltage  drop  in  the  part  AC  of  the  line.  This  will  give  the 
voltage  across  CD,  and  consequently  the  value  of  the  leakage 
current.  After  this,  the  drop  in  r2  is  determined,  and  thus  the 
voltage  across  the  resistance  r  is  found.  Ans.  447.3  volts; 
17.88  amps. 

3.  Resistivity  and  Conductivity.  A  cylindrical  conductor 
can  be  considered  as  a  combination  of  unit  conductors  in  series  and 
in  parallel.  For  instance,  a  wire  12  m.  long  and  having  a  cross- 


CHAP,  i]  DIRECT-CURRENT  CIRCUITS.  7 

section  of  70  sq.  mm.  can  be  regarded  as  composed  of  70  X  12  = 
840  unit  conductors,  each  of  one  square  millimeter  cross-section, 
and  one  meter  long.  These  unit  conductors  are  first  combined 
into  sets  of  70  in  parallel,  and  then  12  sets  are  connected  in  series. 
The  resistance  of  such  a  unit  conductor,  made  of  copper,  and  at 
a  temperature  of  o°  C.,  is  about  .016  ohm.  A  set  of  70  unit 
conductors  in  parallel  has  70  times  less  resistance,  because  the 
current  is  offered  70  paths,  instead  of  one  ;  twelve  sets  connected 
in  series  offer  twelve  times  greater  resistance  than  each  set. 
Therefore  the  resistance  of  the  given  conductor  is  (.016/70)  X 
12  =  .002743  ohm. 

Each  material  is  characterized  by  the  resistance  of  a  unit  con- 
ductor made  out  of  it.     The  resistance  of  a  unit  conductor  is 
called  the  resistivity*  of  the  material  and  is  denoted  by  p.    Thus, 
the  resistance  of  a  conductor  of  a  length  /  and  cross-section  A  is 

r=p-/IA  ........     (8) 

The  numerical  value  of  p  depends  upon  the  units  of  length  and 
resistance  used.  A  unit  conductor  may  have  the  cross-section 
of  one  square  millimeter,  and  be  one  meter,  or  one  kilometer 
long  ;  or  it  may  be  a  centimeter  cube.  In  the  Hnglish  system  it 
may  have  the  cross-section  equal  to  one  circular  mil,  and  be  one 
foot  long,  or  one  thousand  feet  long,  etc.  The  resistance  of  a 
unit  conductor  can  be  expressed  in  ohms,  megohms,  microhms, 
etc.  One  megohm  =  one  million  ohms  ;  one  microhm  =  one 
millionth  part  of  an  ohm.  In  each  case,  the  unit  of  resistance, 
and  the  units  in  which  the  dimensions  of  /and  A  are  expressed 
in  formula  (8),  are  selected  so  as  to  suit  the  convenience  of  the 
user  of  the  formula.  For  the  values  of  p  for  various  materials, 
and  for  the  effect  of  temperature,  see  various  handbooks  and 
pocket-books  for  electrical  engineers. 

In  some  cases  it  is  more  convenient  to  use  the  conductance  of 
the  unit  conductor,  instead  of  its  resistance.  The  conductance 
of  a  unit  conductor  is  called  the  electric  conductivity  of  the 
material,  at  a  certain  temperature  ;  it  is  equal  to  the  reciprocal 
of  its  resistivity.  Denoting  this  conductivity  by  y  we  have  : 
y  =  ijp  •  the  conductance  of  a  conductor  having  dimensions  /  and 
.4  is 

.......     (9) 


The  older  name  is  '  '  specific  resistance.  '  ' 


DIRECT-CURRENT  CIRCUITS.  [CHAP,  i 

PROB.  4.  Resistivity  of  aluminum  is  =  2.66  microhms  per  cu. 
cm.;  what  is  its  conductivity  per  mil- foot?  Solution:  Resist- 
ance of  a  conductor  one  foot  long  and  having  its  cross-section 
equal  to  one  circular  mil  is  2.66  X  icf6  X  197300  X  30.48  =  16 
ohms  ;  where  197300  =  ( 1000/2. 54)2  x  4/73- is  the  conversion  factor 
of  square  centimeters  into  circular  mils  ;  30.48  is  the  ratio  of  one 
foot  to  one  centimeter.  Ans.  Conductivity  of  aluminum  is 
=  1/16  =  .0625  mho.  per  circular  mil  foot. 

4.  Current  Density  and  Electric  Intensity.  The  use  of  the 
coefficients  of  resistivity  and  conductivity  permits  of  the  follow- 
ing interpretation  of  Ohm's  law,  that  is  convenient  in  some  ap- 
plications :  Substitute  the  value  of  g  from  eq.  (9)  into  (6),  and 
divide  the  result  by  A  ;  we  get :  i\A  =  y  •  ejl,  or 

U=yF    .     .     ...  •  -..     .      (10) 
where 

U=i\A      .     ..    .     .    v;   .     .    (n) 
and 

F=e\L-    .     .     .     .•    ...  ,    .     (12) 

The  quantity  U  is  called  current  density,  being  equal  to  the 
current  iper  unit  cross-section  of  the  conductor.  The  quantity  F 
is  called  electric  intensity ;  it  represents  the  voltage  drop  per  unit 
length  of  the  conductor,  or  the  rate  of  change  of  the  voltage 
with  the  length.  Equation  (10)  states  Ohm's  law  in  application 
to  a  unit  conductor,  because  U  is  the  current  flowing  through 
the  unit  conductor,  F'\$>  the  voltage  across  it,  and  y  is  the  con- 
ductance of  the  unit  conductor.  In  this  form,  the  relation  is 
independent  of  the  actual  dimensions  of  the  conductor,  which 
fact  is  of  advantage  in  the  solution  of  some  problems.* 

PROB.  5.  What  is  the  voltage  drop  per  kilometer  in  a  copper 
wire,  having  a  cross-section  of  70  sq.  mm.,  and  carrying  a  current 


*  The  treatment  in  Sections  4  and  5  is  intended  to  prepare  the  student's 
mind  to  a  reasoning  in  terms  of  the  quantities  which  are  of  importance  in 
electrostatics  and  in  magnetism.  Eq.  (10)  has  its  perfect  analogs  in  the 
electrostatic  and  in  the  magnetic  circuits  :  to  the  current  density  correspond 
there  the  dielectric  flux  density  and  the  magnetic  flux  density.  The  electric 
intensity  is  analogous  to  the  electrostatic  intensity  and  to  the  magnetic 
intensity ;  while  conductivity  finds  its  analogs  in  permittivity  and  in 
permeability.  See  Chapter  VI  of  this  pamphlet,  and  the  pamphlet  on  ''The 
Magnetic  Circuit. ' ' 


CHAP,  i]  DIRECT-CURRENT  CIRCUITS.  9 

of  150  amperes?  Solution:  U '=  150/70=  2.143  amps  per  sq. 
mm.  Conductivity  of  copper  is  y  =  57  mhos,  for  a  unit  con- 
ductor of  one  sq.  mm.  cross- section,  and  one  meter  long.  There- 
fore, the  electric  intensity,  or  the  voltage  drop  per  meter  of 
length,  is,  according  to  formula  (10),  F=  2.143/57  =  -°376  volt, 
per  meter  length.  Ans.  37.6  volt. /km. 

PROB.  6.  What  is  the  expression  for  power  converted  into 
heat  in  a  unit  conductor?  Ans. 

Power  =  F-  U  =  U2P  =  U*ly  =  y  F\     .     .      (13) 

PROB.  7.  What  is  the  amount  of  power  lost  in  the  conductor 
considered  in  problem  5?  Ans.  (.0376  X  2.143)  X  70  X  1000  = 
5640  watts. 

PROB.  8.  The  space  available  for  a  rectangular  field-coil,  on 
the  frame  of  a  generator,  is  16  X  12  cm.  for  the  inside  dimen- 
sions, and  28  X  24  cm.  for  the  outside  dimensions  ;  the  available 
height  is  15  cm.  What  current  density  can  be  allowed  in  such 
a  coil,  if  12  sq.  cm.  of  exposed  surface  are  required  per  watt  loss, 
in  order  that  the  temperature  of  the  coil  shall  not  exceed  the 
safe  limit?  The  space  factor  of  the  coil  is  .55,  in  other  words, 
55  per  cent  of  the  gross  space  is  occupied  by  copper,  the  rest 
being  taken  by  the  air  spaces  and  the  insulation.  Solution : 
Exposed  surface  =  2(28  -f  24)  X  15  =  1560  sq.  cm.;  therefore, 
130  watts  loss  can  be  allowed  in  the  coil.  With  a  space  factor 
of  .55  the  useful  cross-section  of  copper  is  6  X  15  X  .55  =  49.5 
sq.  cm.;  the  average  length  of  one  turn  is  =  2(22  +  18)  =  80 
cm.  Therefore,  the  coil  contains  4950  X  .8  =  3960  unit  con- 
ductors, each  one  meter  long  and  one  square  millimeter  in  cross- 
section.  Permissible  loss  per  unit  conductor  is  130/3960=  .0328 
watts.  Equation  (13)  gives  then 


£/=  x/  .0328  X  57  =  1.37  amps,  per  sq.  mm. 

This  result  is  independent  of  the  size  of  the  wire,  as  long  as 
the  space  factor  remains  approximately  the  same.  This  example 
shows  the  convenience  of  the  concept  of  unit  conductor. 

5.  Conductors  of  Variable  Cross-section.  When  the  cross- 
section  of  a  conductor  varies  along  its  length,  the  voltage  drop 
per  unit  length  and  the  current  density  are  also  variable.  In 
places  where  the  cross-section  of  the  conductor  is  comparatively 


io  DIRECT-CURRENT  CIRCUITS.  [CHAP,  i 

large,  the  resistance  per  unit  length  is  comparatively  small,  and 
vice  versa.  The  current  density,  and  the  electric  intensity,  also 
vary  from  cross- section  to  cross- section.  The  relation  between 
the  variable  intensity  /'along  the  conductor,  and  the  total  voltage 
e  at  its  terminals  is  no  more  expressed  by  the  simple  relation  (12), 
applicable  to  the  whole  conductor.  Relation  (12)  must  be  now 
written  for  an  infinitesimal  length  ds  of  the  conductor,  because 
FCSLU  be  considered  constant  only  for  an  infinitesimal  length. 
The  definition  of  F  remains  the  same,  namely,  F  is  the  rate  of 
variation  of  voltage  per  unit  length  of  the  conductor.  Thus,  we 
have : 

F—de\ds,  or  de  =  F-ds     ....     (14) 

where  de  is  the  voltage  drop  in  the  element  ds  of  the  con- 
ductor. Total  voltage  e  at  the  terminals  of  the  conductor  is 
equal  to  the  sum  of  these  infinitesimal  drops,  or 


e  =  J     F-ds .      (15) 

o 

Eq.  (15)  is  usually  expressed  in  words  by  saying  that  voltage  is 
the  line  integral  of  electric  intensity. 

A  clear  understanding  of  the  relations  (14)  and  (15)  is  of 
paramount  importance  in  the  study  of  electro-static  and  magnetic 
phenomena  ;  therefore  eqs.  (14)  and  (15)  are  further  expounded 
below,  by  resorting  to  the  analogies  used  before.  In  the  case  of 
flow  of  heat,  /^corresponds  to  the  rate  of  change  in  temperature 
per  unit  length  of  the  rod,  while  e  represents  the  total  difference 
of  temperatures  at  the  ends  of  the  rod.  Eq.  (15)  merely  ex- 
presses, that  by  taking  the  rate  at  a  certain  point  and  multiplying 
it  by  a  very  short  element  of  the  length  of  the  rod,  the  actual 
difference  of  temperatures  at  the  ends  of  this  element  is  obtained. 
Thus,  let,  for  instance,  the  drop  in  temperature  at  some  point  of 
the  rod  vary  at  a  rate  of  2.5  degrees  Centigrade  per  meter 
length.  Then  the  actual  drop  upon  a  very  short  element,  say 
o.i  mm.,  is  2.5  x  .0001  =  .00025°  C.  The  element  of  length 
must  be  small,  because,  by  supposition,  the  cross-section  of  the 
rod  is  variable,  and  the  rate  of  drop  is  consequently  variable. 
For  a  short  length  the  variable  quantities  can  be  assumed 
constant,  or,  more  correctly,  average  values  can  be  used. 


CHAP,  i  ]  DIRECT-CURRENT  CIRCUITS.  n 

Similarly,  in  a  pipe  of  variable  cross-section  the  rate  of  loss  of 
head  per  unit  length  is  variable,  so  that  it  is  only  possible  to 
speak  of  this  rate  /'"at  a  point.  Total  loss  of  pressure,  or  the  head 
e,  is  obtained  by  summing  up  the  small  losses  of  head  on  infini- 
tesimal elements  of  the  pipe.  The  loss  of  pressure  for  a  length  ds 
is  Fds\  total  head  e  is  the  integral  of  this  expression  over  the 
total  length  of  the  pipe.  This  is  expressed  mathematically  by 

eq.  (15). 

The  following  examples  illustrate  some   methods  of  dealing 

with  conductors  of  non-uniform  cross-section. 

PROB.   9.    Calculate   the  resistance  of  a  cylindrical  layer  of 

mercury  MM  (Fig.  2)  of  height  h  =  5  cm.,  between  two  con- 
centric cylindrical  electrodes  Al  and  Az, 
the  radii  of  the  surfaces  of  contact  being 
#!  =  10  cm.  and  a2  —  18  cm.  Resistivity 
of  mercury  is  p  =  95  microhms  per  cu. 
cm.  Solution  :  Take  an  infinitesimal 
layer  of  the  mercury,  between  the  radii 
x  and  x  -f-  dx  ;  the  resistance  of  this 
layer,  according  to  eq.  (8),  is 

dr  =  p  •  dx\2-jrxh. 
FIG.  2.     Flow  of  current 

between  two  concentric    The  resistances  of   all  the   infinitesimal 
electrodes.  concentric  layers  are  in  series,   so  that 

integrating  this  expression  between  the  limits  al  and  a2  gives 


Ln  ajal  ......       (16) 

Ans.  r=  1.775  microhms.* 

PROB.  10.  When  a  current  z  =  10.000  amps,  flows  through  the 
mercury  in  the  preceding  problem,  what  is  the  amount  of  heat 
generated  per  second  per  cu.  cm.  of  mercury,  at  both  electrodes? 
Solution  :  Current  density  at  the  inner  electrode  is  io.ooo/(2?r  x 
I0  x  5)  =  31.  8  amps,  per  sq.  cm.  From  eq.  (13)  we  find  there- 
fore :  loss  of  power  =  (31.  8)2  X  95  X  io~6=  .0958  watts  per 
cu.  cm.  The  heat  loss  at  the  outer  electrode  is  .096  X  (io/i8)2  = 
.0295  watts  per  cu.  cm. 

PROB.  n.  What  is  the  curve  of  electric  intensity  in  the 
arrangement  given  in  problems  9  and  10,  and  what  are  the  limit- 


*  Electrostatic  capacity  between  two  concentric  cylinders  is  obtained  by 
means  of  an  identical  reasoning  ;  see  problem  121,  in  Chapter  VI. 


12  DIRECT-CURRENT  CIRCUITS.  [CHAP,  i 

ing  values  of  F?  Solution  :  From  eq.  (10),  substituting  i/p  for 
y,  we  have  :  F  =  p  U.  But  in  our  case  U  is  =  i\2irxh  ;  hence, 
F=pi\2Trh>  ijx.  This  is  the  equation  of  an  equilateral  hyper- 
bola, if  F  be  plotted  to  x  as  abscissae.  The  result  could  have 
been  forseen,  because  the  cross-section  of  the  path  of  the  current 
varies  inversely  as  the  distance  from  the  center.  Therefore,  the 
current  density  and  the  electric  intensity  must  vary  inversely  as 
the  distance  from  the  center.  The  value  of  Fat  the  inner  elec- 
trode is  3.02  milli- volts  per  cm.  length ;  at  the  outer  electrode 
F==  1.677  milli- volts  per  cm. 

PROB.  12.  A  lead-covered  cable,  consisting  of  a  solid  circular 
conductor  of  A  sq.  mm.  in  cross- section,  is  insulated  with  a  layer 
of  rubber  a  mm.  thick,  between  the  conductor  and  the  sheathing. 
What  is  the  insulation  resistance  of  /  kilometers  of  such  a  cable, 
if  the  resistivity  of  rubber  is  y  megohms  per  centimeter  cube? 
Ans.  p  X  io~5/27r/'  L,n(i  +  i.fjtj2aj\^A)  megohms,  according  to 
eq.  (16). 

PROB.  13.  Show  that  by  increasing  the  thickness  of  insulation 
in  the  preceding  problem  twice,  the  insulation  resistance  is 
increased  less  than  twice. 

PROB.  14.  Current  is  flowing  across  a  hemispherical  shell  of 
metal  along  its  radial  lines.  Express  the  resistance  of  the  shell 
as  a  function  of  its  radii  bl  and  b^,  and  the  conductivity  c  of  the 
material.  Ans.  (bz  — b^)l(2ircblbz). 

PROB.  15.  A  non-linear  irregular  conductor,  made  of  homo- 
geneous material,  has  the  current  density  U  and  the  electric 
intensity  F  varying  from  point  to  point  in  magnitude  and  direc- 
tion. What  is  the  general  expression  for  the  power  lost  by 
Joulean  heat?  Ans.  According  to  eq.  (13), 

Power=   CF-  Udv=l§  U*dv  =  y  §F*dv    .    .(17) 

where  dv  is  the  element  of  volume  to  which  Fznd  U  refer,  and  the 
integration  is  extended  over  the  whole  volume  of  the  conductor. 
The  volume  dv  must  be  taken  as  a  cylinder  or  parallelepiped, 
the  length  of  which  is  in  the  direction  of  flow  of  current,  and 
the  cross-section  of  which  is  perpenducular  to  this  flow. 


CHAPTER  II. 

REPRESENTATION  OF  ALTERNATING  CURRENTS 

AND  VOLTAGES  BY  SINE  WAVES  AND 

BY  VECTORS. 

6.  Sinusoidal  Voltages  and  Currents.  A  large  proportion  of 
electric  power  used  for  lighting,  industrial  purposes,  and  traction, 
is  generated  in  the  form  of  alternating  currents.  Some  of  the 
advantages  of  the  alternating  current  over  the  direct  current  are  : 
( i )  Alternating  power  can  be  easily  converted  into  power  at  a 
higher  or  at  a  lower  voltage,  thus  making  possible  the  transmis- 
sion of  power  over  long  distances.  (2)  The  generation  of  alter- 
nating currents  is  simpler  than  that  of  direct  currents.  The 
latter  require  the  use  of  a  commutator,  which  is  liable  to  give 
troubles  in  operation.  (3)  By  combining  two  or  three  alternat- 
ing-current circuits  into  the  so-called  polyphase  systems  it  is 
possible  to  convert  electric  power  into  mechanical  by  motors  of 
simple  and  rugged  construction  (induction  motors  and  synchro- 
nous motors). 


FIG.  3.     An  alternating  current  represented  by  a  sine  wave. 

Alternating  voltages,  generated  by  commercial  alternators, 
are  more  or  less  irregular  in  shape,  but  for  most  engineering 
calculations  it  is  accurate  enough  to  assume  them  to  vary  with 
the  time  according  to  the  sine  law  (Fig.  3).  This  assumption 


14  SINE  WAVES  AND  VECTORS,  [CHAP.  2 

simplifies  the  theory  and  calculations  greatly,  and  at  the  same 
time  gives  comparable  results,  the  same  referring  to  a  standard 
shape  of  the  voltage  and  current  curves,  instead  of  particular 
forms  in  a  specific  problem.  Moreover,  if  the  curve  of  a  voltage 
or  current  differs  greatly  from  the  sine-wave,  it  can  be  resolved 
into  a  series  of  sine-waves  of  different  frequencies,  so  that  even 
then  the  sine-wave  remains  the  fundamental  form  (see  Kara- 
petofl's  Experimental  Electrical  Engineering,  under  "wave 
analysis")-  Fig.  3  shows  the  well-known  construction  of  a 
sine-  wave,  instantaneous  values*  of  the  voltage  or  the  current 
being  represented  as  ordinates,  against  time  as  abscissae.  Instead 
of  actual  time  in  seconds,  the  curve  can  be  plotted  against  some 
other  quantity  proportional  to  time,  for  instance,  fractions  of  a 
complete  period,  or  angular  positions  of  the  field  pole  of  the 
alternator  with  respect  to  a  conductor  in  which  the  electro-motive 
force  under  consideration  is  induced. 

Draw  a  circle  with  the  radius  equal  to  the  maximum  value  of 
alternating  current  (or  voltage)  the  wave  of  which  is  desired  to 
construct.  Divide  the  circle  into  a  certain  number  of  equal  or 
unequal  parts,  such  as  ad,  be,  etc.,  and  mark  on  the  axis  of 
abscissae  points  a',  b'  ',  c'  ',  etc.  corresponding  to  the  points  of 
division  on  the  circle.  That  is  to  say,  a'c'  is  either  equal  or  pro- 
portional to  ac  ;  in  general,  a'c'  represents,  to  a  certain  selected 
scale,  the  central  angle  x  corresponding  to  the  arc  ac.  The 
length  a'm'  represents  to  the  same  scale  the  angle  of  360  degrees, 
or  also  the  time  of  one  complete  cycle  of  the  wave.  The  point 
£*"  on  the  curve  is  obtained  by  transferring  'the  ordinate  cc"  of 
the  circle  to  the  corresponding  abscissa  a'c'  .  The  name  sine- 
wave  is  derived  from  the  fact  that  these  ordinates  are  proportional 
to  the  sines  of  the  corresponding  angles,  such  as  x. 

The  equation  of  the  curve  is  obtained  by  expressing  this 
property  analytically.  Let  the  maximum  value  of  the  current, 
which  is  also  equal  to  the  radius  of  the  circle,  be  denoted  by  /m  ; 
we  have  then  from  the  triangle  Occ" 

......      (18) 


where  the  ordinate  i  =  cc"  =  c'c'"  represents  the  instantaneous 
value  of  the  alternating  current,  at  the  moment  of  time  corre- 


CHAP.  2]  SINE   WAVES  AND  VECTORS.  15 

spending  to  the  angle  x.  The  variable  angle  x  is  proportional 
to  time,  because  the  radius  Oc  which  generates  the  sine  wave 
revolves  at  a  uniform  speed.  Let  time  /  be  counted  from  the 
position  Oa  of  this  radius,  and  let  T=  a'm'  be  the  interval  of 
time  necessary  to  complete  one  revolution  of  the  radius,  or  the 
time  of  one  complete  cycle  of  the  alternating  wave.  When  time 
/  =  o,  x  =  o ;  when  t  =  T,  x  =  2?r.  Therefore,  in  general,  x  = 
2irt\T\  because  this  expression  satisfies  the  foregoing  conditions. 
We  have  then  : 

i  =  /m  sin  2trtl  T.     .     .     .     .     .     .    (19) 

For  the  values  of  /  =  o,  TJ2,  T,  37)2,  etc.,  z  =  o,  as  it  ought 
to  be,  because  at  these  moments  the  current  changes  from  its 
positive  to  negative  values  or  vice  versa.  At  /  =  7)4,  3  7)4,  5  7)4, 
etc. ,  we  have  i  =  ±  7m  ;  at  these  moments  the  current  reaches  its 
positive  and  negative  maxima.  Eq.  (18)  is  used  when  the  sine- 
wave  is  plotted  against  values  of  angle  x  as  abscissae.  Eq.  (19) 
gives  the  same  curve  referred  to  time  t  as  abscissae. 

In  practice,  the  rapidity  with  which  currents  and  voltages 
alternate  is  not  denoted  by  the  fraction  of  a  second  T  during 
which  a  cycle  is  completed,  but,  in  a  more  convenient  manner, 
by  the  number  of  complete  cycles  in  a  second.  Thus,  instead 
of  saying  that  an  alternator  generates  current  which  completes 
a  cycle  within  i/6oth  of  a  second,  it  is  customary  to  say  that  the 
frequency  of  the  current  is  60  cycles  per  second.  Denoting  the 
frequency  in  cycles  per  second  by /we  have 

/=i/7* (20) 

and  consequently 

i  —  7m sin  2-nft. (21) 

This  is  the  usual  expression  for  an  alternating  current  of 
frequency /periods  per  second.  Analogously,  for  an  alternating 
voltage  we  have : 

e  —  E^ sin  2-nft (22) 

where  Em  is  the  maximum  instantaneous  value,  also  called  the 
amplitude,  and  e  is  the  instantaneous  value  at  the  moment  t. 


1 6  SINE  WAVES  AND  VECTORS.  [CHAP.  2 

In  numerical  calculations,  and  in  drawing  sine  waves,  the 
values  of  ordinates  for  various  values  of  x  or  t  are  obtained  either 
graphically,  as  shown  in  Fig.  3,  or  from  tables  of  sines.  For 
approximate  calculations,  values  of  sines  can  be  taken  from  a 
slide-rule.  In  the  problems  16  to  20,  that  follow,  the  student  is 
advised  to  become  familiar  with  all  the  three  methods  of  getting 
values  of  sines. 

PROB.  16.  An  alternating  current  fluctuates  according  to  the 
sine  law  between  the  values  ±  7m  =  75  amps.,  making  n  =  6000 
alternations  per  minute  (3000  positive  and  as  many  negative 
ones).  Draw  a  curve  of  instantaneous  values  of  this  current  ; 
mark  on  the  axis  of  abscissae  :  time  t  in  thousandths  of  a  second, 
angles  x  in  degrees,  and  the  same  angles  in  radians. 

PROB.  17.  What  is  the  frequency  of  the  current  in  the  preced- 
ing problem,  in  cycles  per  second?  Ans.  /=  50. 

PROB.  1 8.  Plot  on  the  same  curve  sheet  with  the  curve  obtained 
in  problem  16  the  sine- wave  of  a  current  the  frequency  of  which 
is  three  times  higher,  the  amplitude  7m  —  52  amps.  The  curve 
is  to  be  at  its  maximum  when  the  first  curve  is  at  a  maximum. 

PROB.  19.  Supplement  the  preceding  curves  by  one  the 
frequency  of  which  /=  50  cycles  per  second,  the  amplitude 
7m'  =  63  amps.,  and  which  reaches  its  positive  maxima  at 
moments  in  which  the  first  curve  passes  through  zero.  Show 
that  with  these  specifications  two  distinct  curves  can  be  drawn. 

PROB.  20.  Draw  on  the  same  curve  sheet  with  the  preceding 
curves  a  sine-wave  representing  an  alternating  current  of  the 
same  frequency,  f=  50  periods  per  second,  and  which  is  deter- 
mined by  the  following  conditions:  Its  amplitude  /m"  =  120 
amps. ;  it  passes  through  zero  and  begins  to  increase  in  the 
positive  direction  1/600  of  a  second  later  than  the  first  curve. 
Thus,  it  passes  through  zero  at  an  angular  distance,  <f>  —  30 
degrees,  to  the  right  of  the  first  curve.  In  the  usual  engineering 
language,  the  first  current  is  said  to  lead  the  second  by  30  degrees, 
or  the  second  to  lag  behind  the  first  by  the  same  amount.  The 
angle  <£  is  called  the  angle  of  phase  difference  between  the  two 
currents. 


CHAP.  2]  SINE  WAVES  AND  VECTORS.  17 

PROB.  2.1.  The  current  mentioned  in  problem  16  is  generated 
by  a  twelve-pole  alternator,  that  in  problem  18  by  a  fourteen-pole 
machine.  At  what  speeds  must  these  machines  be  driven  in 
order  to  give  the  required  frequencies?  Ans.  500  and  1285  r.p.m. 

PROB.  22.  Express  the  currents  given  in  problems  16  to  20  by 
equations  of  the  form  of  equation  (18).  Ans.  i  =  75  sin  x ; 
i  =  —  52  sin  3-* »'  z  =  ±  63  cos  x ;  i  =  120  sin  (x  —  30°). 

PROB.  23.  The  angle  x  in  the  answers  to  the  preceding  problem 
is  expressed  in  degrees ;  rewrite  the  equations  so  as  to  have  x 
expressed  in  radians,  and  in  fractions  of  a  cycle.  Also  represent 
the  currents  as  functions  of  time  t. 

PROB.  24.  Express  by  equations  similar  to  equation  (22)  the 
following  sinusoidal  voltages  of  frequency/:  (a)  Amplitude  Em 
volts,  (b)  Amplitude  E^  volts,  lagging  a  degrees  with  respect 
to  the  first  curve,  (c)  Amplitude  £m"  volts,  leading  the  second 
curve  by  <£  radians,  (d)  Amplitude  En'"  volts,  lagging  i/wth  of 
a  cycle  with  respect  to  curve  (a). 

PROB.  25.  The  voltages  required  in  the  preceding  problem  are 
induced  by  four  identical  alternators,  having  p  poles  each,  and 
coupled  together.  By  what  geometrical  angles  must  the  revolv- 
ing or  the  stationary  parts  be  displaced  in  order  to  give  the 
required  differences  in  phase  ? 

7.  Representation  of  a  Sine-wave  by  a  Vector.  The  fore- 
going problems  make  clear  that  all  sine- waves  are  different  from 
one  another  in  three  respects  only,  namely  :  ( i )  In  amplitude  ; 
(2)  in  frequency;  (3)  in  relative  phase  position.  In  most 
practical  cases,  all  the  currents  and  voltages  entering  into  a 
problem  are  of  the  same  frequency,  so  that  they  differ  from  each 
other  solely  in  their  amplitudes  and  phase  positions.  It  is  not 
necessary  in  such  cases  actually  to  draw  sine-waves,  or  even  to 
write  their  equations :  It  is  sufficient  to  indicate  the  radii  generat- 
ing these  curves  (Fig.  4),  in  their  true  magnitudes  and  relative 
positions.  The  rotating  radius,  at  different  instants,  gives  by  its 
vertical  projection  to  scale  the  magnitude  of  the  alternating  cur- 
rent or  voltage  at  that  instant.  The  initial  position  of  the  radius, 
at  the  instant  we  choose  to  begin  counting  time,  will  therefore 
represent  graphically  the  alternating  quantity.  These  radii  can 
be  either  actually  drawn  and  problems  solved  graphically,  or  the 


i8 


SINE  WA  VES  AND  VECTORS. 


[CHAP.  2 


radii  can  be  expressed  by  their  projections  on  any  pair  of  co- 
ordinate axes,  and  treated  analytically.  This  chapter  is  devoted 
to  the  graphical  solution  of  alternating-current  problems  ;  the 
analytical  treatment  by  projections  is  taken  up  in  the  following 
chapters. 

Fig.  4  shows  the  two  sine- waves  given  in  problems  16  and  20, 
and  the  two  radii  generating  these  two  waves.     It  will  be  seen 


FIG.  4.     Two  alternating  currents  displaced  in  phase  by  30  degrees. 

that  the  angle  between  the  radii  is  30  degrees,  corresponding  to 
the  phase  displacement  between  the  two  waves.  In  Fig.  3,  the 
curve  is  generated  by  the  radius  revolving  counter-clock-wise  ; 
the  radius  representing  the  second  curve  in  Fig.  4  is  shown 
lagging  behind  the  first  radius  in  accordance  with  this  assump- 
tion. The  absolute  position  of  the  radii  is  irrelevant,  because 
they  are  revolving  all  the  time.  It  is  their  relative  position 
which  is  permanent  and  which  determines  the  relative  position  of 
the  sine-waves.  The  moment,  from  which  time  is  counted,  is 
arbitrary  in  most  problems  ;  hence,  one  of  the  radii  can  be  drawn 
in  any  desired  position.  Then,  all  other  radii  in  the  same  problem 
become  definite  by  their  phase  displacements  with  respect  to  this 
' '  reference  ' '  radius. 

It  must  be  clearly  understood  that  the  foregoing  representation 
by  vectors  is  true  only  when  all  the  vectors  are  revolving  at  the 
same  speed,  that  is  to  say,  when  we  are  dealing  with  alternating 
quantities  of  the  same  frequency.  When,  however,  currents  and 
voltages  of  different  frequencies  enter  into  a  problem,  the  angle 
between  the  vectors  varies  all  the  time,  and  it  is  necessary  to 
introduce  an  arbitrary  zero  of  time  for  reference.  In  general, 
graphical  solution  is  unsuitable  for  such  problems. 


CHAP.  2.]  SINE  WAVES  AND  VECTORS.  19 

In  mathematics  and  physics,  a  quantity  which  has  not  only  a 
magnitude,  but  also  a  definite  direction  in  space  or  in  a  plane,  is 
called  a  vector.  Thus,  for  instance,  a  force  in  mechanics  is  a  vector 
quantity,  while  a  volume  is  not  a  vector  quantity.  The  radii 
which  represent  sine-waves  have  both  magnitude  and  direc- 
tion in  a  plane.  It  is  proper,  therefore,  to  call  them  vectors. 
While  the  direction  of  the  first  vector  is  usually  arbitrary,  once 
it  is  selected,  the  directions  of  all  other  radii  become  altogether 
definite,  so  that  with  this  limitation  radii  in  alternating-current 
problems  have  definite  directions  and  can  be  called  vectors. 
While  they  must  be  imagined  as  revolving  when  generating  the 
corresponding  sine-waves,  yet  they  revolve  as  a  system,  all  to- 
gether. The  required  relations  always  depend  upon  the  relative 
positions  of  the  radii,  so  that  the  fact  that  they  are  revolving  can 
be  altogether  disregarded,  and  the  radii  considered  as  ordinary 
stationary  vectors. 

PROB.  26.  Draw  in  Fig.  4  the  vectors  of  the  sine-waves  of 
problem  19. 

PROB.  27.  A  single-phase  alternator  has  a  terminal  voltage 
such  that  its  maximum  instantaneous  value  Em  is  equal  to  16 
kilovolts.  The  maximum  value  of  the  current  supplied  by  the 
machine  is  7m  =  325  amps.  The  character  of  the  load  is  such 
that  the  current  wave  lags  behind  the  voltage  wave  by  an  angle 
<£  =  37  degrees.  Assuming  both  the  voltage  and  the  current  to 
vary  according  to  the  sine  law,  represent  the  foregoing  conditions 
by  two  vectors. 

PROB.  28.  Draw  a  vector  diagram  showing  the  phase  voltages 
(star  voltages)  of  a  25  cycle  three-phase  system,*  the  amplitude 
of  each  voltage  being  equal  to  7235  volts,  and  displaced  in  phase 
by  120  degrees  with  respect  to  the  other  two  voltages.  The 
current  in  the  first  phase  is  equal  to  30  amps,  and  lags  behind 
the  corresponding  phase  voltage  by  i/6th  of  a  cycle.  The  current 
in  the  second  phase  is  47  amps,  and  leads  its  voltage  by  18  de- 
grees. The  current  in  the  third  phase  is  equal  to  72  amps.,  and 
lags  behind  the  corresponding  phase  voltage  by  .004  of  a  second. 


*  See  the  chapter  on  the  ' '  Electrical  Relations  in  Polyphase  Systems ' '  in 
the  author's  "Experimental  Electrical  Engineering." 


20  SINE  WA  VES  AND  VECTORS,  [CHAP.  2. 

NOTE.  The  student  is  supposed  to  draw  the  vectors  in  problems 
26  to  28  to  represent  the  amplitudes  of  the  corresponding  alternat- 
ing waves.  It  is  customary  in  practice  to  draw  vectors  of  the 
effective  values  of  voltages  and  currents,  and  not  of  the  ampli- 
tudes ;  for  sine- waves  the  effective  value  is  equal  to  the  amplitude 
divided  by  ^2  (see  Chapter  III).  The  advantage  of  using 
effective  values,  instead  of  amplitudes,  is  explained  at  the  end  of 
Sec.  9  below.  The  difference  is  not  essential  for  our  present  pur- 
poses. The  use  of  effective  values  would  merely  change  the  scale 
to  which  the  vectors  are  drawn,  which  scale  is  arbitrary  anyway. 

8.  Addition  and  Subtraction  of  Vectors.  There  are  many 
occasions  when  alternating  currents  and  voltages  have  to  be 
added  together,  and  subtracted  from  one  another.  For  instance, 
when  two  or  more  alternators  are  working  in  parallel  the  total 
current  delivered  to  the  bus-bars  of  the  station  is  equal  to  the 
sum  of  the  currents  supplied  by  each  machine.  Or,  to  find  the 
voltage  at  the  receiving  end  of  a  transmission  line  the  voltage 
drop  in  the  line  is  subtracted  from  the  generator  voltage.  When 
the  component  quantities  vary  according  to  the  sine  law  the 
resultant  quantity  is  also  a  sine  curve.  This  curve  can  be  found 
graphically,  by  adding  the  given  curves  point  by  point,  or  by 
adding  their  equations  ;  also  by  adding  the  vectors  of  these 
curves.  All  these  metods  are  explained  and  illustrated  below. 

PROB.  29.  The  currents  given  in  problems  16  and  20  are 
generated  by  two  alternators  working  in  parallel.  Find  the 
curve  of  the  total  current  by  adding  the  ordinates  of  the  curves 
point  by  point.  Measure  the  amplitude  and  the  phase  position 
of  the  resultant  current,  and  write  down  its  equation.  Ans.  Ap- 
proximately 189  sin  (x  —  1 8°  30'),  in  amperes. 

It  can  be  proved  analytically  that  the  sum  of  two  sine-waves 
of  the  same  frequency  is  also  a  sine  wave,  of  the  same  frequency. 
Therefore,  it  is  not  necessary  to  add  sine  waves  point  by  point ; 
the  equation  of  the  resultant  wave  can  be  derived  directly  from 
the  equations  of  the  component  curves.  Let  one  wave  be  repre- 
sented by  eq.  (18);  the  other  wave,  which  we  shall  assume  to 
lead  the  first  by  an  angle  <£,  can  be  represented  by  the  equation 
i'  =  7m'  sin  (^  -f  <£).  The  sum  of  their  ordinates  is 

i  -f  i'  =  (/m  -f  /m'  cos  <£)  sin  x  -f  /m'  sin  <£  •  cos  xy 


CHAP.  2.]  SINE   WAVES  AND  VECTORS.  21 

where  the  constant  terms  multiplied  by  sin  x  and  by  cos  x  are 
collected  together.  The  form  of  the  equation  suggests  the  sine 
of  the  sum  of  two  angles.  We  denote  therefore 


cos  <£  =    „  cos 

sin  <*>  =  /    sin  * 


and  obtain 


where  the  subscript  '  '  equivalent  '  '  refers  to  the  amplitude  and 
the  phase  of  the  resultant  wave.  The  last  equation  shows  that 
the  resultant  current  zeq  varies  according  to  the  sine  law,  and  has 
the  same  frequency  as  the  component  curves,  because  the  variable 
angle  x  =  2irft  is  the  same  as  in  the  component  waves. 

PROB.  30.  Find  analytically  the  equation  of  the  resultant  wave, 
required  in  problem  29.  Solution  :  The  equations  of  the  given 
currents  are  :  75  sin  x  and  120  sin  (x  —  30°),  so  that  eqs.  (23) 
become  : 

/eqcos^eq  =  75  +  120  x  |-x/3=  178.  92  amps.;)  , 

7^  sin  ^  =  —  120  x  i/2  =  —  60  amps.  J  ' 

Dividing  the  second  equation  by  the  first,  gives  :  tan  ^  =  — 
•  3353,  or  4%  =  —  l8°  32'.  Consequently,  7^  =  -  6o/sin  ^  = 
188.8  amps.  Ans.  188.8  sin  (x  —  i8°32'),  in  amperes. 

PROB.  31.  vSolve  the  preceding  problem  without  the  use  of  eqs. 
(23),  simply  on  the  basis  of  the  theorem  proven  above  that  the 
sum  or  the  difference  of  two  sine  waves  is  also  a  sine  wave. 
Solution  : 

/eqSin  (x  -f  <£eq)  =  75  sin  x  +  120  sin  (x  —  30°). 

This  equation  is  true  for  any  instant,  or  for  any  value  of  x.  It 
contains  two  unknown  quantities,  the  amplitude  and  the  phase 
position  of  the  resultant  curve.  It  is  necessary,  therefore,  to 
apply  this  equation  to  two  particular  moments  of  time,  in  order 
to  obtain  two  equations  with  two  unknown  quantities.  It  is 
most  convenient  in  this  particular  case  to  choose  x  =  7r/2  and 
x  —  o.  Substituting  these  values,  eqs.  (233)  are  obtained.  This 
method  is  preferable  in  the  solution  of  practical  problems,  because 
it  is  not  necessary  to  remember  eqs.  (23). 


22 


SINE  WAVES  AND  VECTORS. 


[CHAP.  2. 


The  foregoing  examples  show  that  adding  alternating  currents 
either  point  by  point,  or  analytically,  is  a  somewhat  tedious 
operation.  It  can  be  dispensed  with  because  of  the  fact  that  the 
resultant  current  is  always  sinusoidal,  and  therefore  can  also  be 
represented  by  a  vector.  The  problem  is  reduced  simply  to  find- 
ing the  vector  of  the  resultant  wave,  knowing  the  vectors  of  the 
component  waves  in  their  magnitude  and  position.  Any  ordinate 
of  the  resultant  wave  must  be  equal  to  the  sum  of  the  correspond- 


i 

FIG.  5.     Addition  of  vectors. 

ing  ordinates  of  the  component  waves.  Hence,  the  vector  of  the 
resultant  wave  must  satisfy  the  condition  that  its  projection  on 
the  K-axis  (Fig.  5)  be  equal  to  the  sum  of  the  projections  of  the 
component  vectors  on  the  same  axis.  This  condition  must  be 
fulfilled  at  all  instants  of  time,  that  is  to  say,  when  the  three 
vectors  are  rotated.  To  satisfy  this  requirement  the  resultant 
vector  must  be  the  closing  side  of  the  triangle  constructed  on 
the  other  two  vectors. 

L,et  OA  and  OB  be  the  given  vectors  to  be  added  together. 
From  the  end  B  of  the  vector  OB  draw  a  line  BC  equal  and 
parallel  to  OA.  Connecting  O  to  C  gives  the  resultant  vector 
OC,  in  magnitude  and  position.  It  is  easy  to  see  from  the  figure 
that  the  projection  of  OC  on  the  K-axis  is  equal  to  the  sum  of  the 
projections  of  OB  and  BC  upon  the  same  axis.  But  BCis  equal 
and  parallel  to  OA,  so  that  the  projection  of  OC  on  the  vertical 
axis  is  equal  to  the  sum  of  the  projections  of  the  given  vectors 
on  the  same  axis.  This  construction  holds  true  for  any  instant. 
By  drawing  AC,  the  parallelogram  OBCA  is  completed,  and  it 


CHAP.  2.] 


SINE   WA  VES  AND  VECTORS. 


23 


will  be  seen  that  the  construction  is  identical  with  that  of  finding 
the  resultant  of  two  forces  in  mechanics.  However,  in  practical 
applications  it  is  not  necessary  to  complete  the  parallelogram, 
because  the  resultant  is  perfectly  determined  from  the  triangle 
OBC.  The  resultant  of  two  vectors  obtained  in  this  way  is  called 
their  geometric  sum. 

If  the  triangle  of  the  vectors  were  not  closed,  the  condition  of 
the   sum   of   projections  might  be  satisfied   for  one  particular 
instant  of  the  cycle,  but  not  be  satisfied  for  any  other  instant. 
Thus,  for  instance,  assuming  line  OC  to  be  the  resultant  vector, 
we  see  that  its  projection  on  axis  OYis  equal  to  the  sum  of  the 
projections  of  OB  and  BC  on  the  same  axis  for  the  given  moment 
of  time  ;  but  the  condition  is  not  fulfilled  when  the  figure  rotates. 
The  rule  for  subtraction  of  vectors  follows  immediately  from 
the  preceding  rule,  because  to  subtract  a  vector  means  to  add  a 
vector  with  the  opposite  sign.     Thus,  let 
it  be  required  to  subtract  vector  OA  from 
OB   (Fig.    6)  ;    this   means  to   subtract 
the  voltage  wave  represented  by  OA  from 
that  represented  by  OB.     From  the  end 
B  of  OB  draw  vector  BC  equal  and  oppo- 
site to  OA.  The  resultant,  OC,  represents 
the  difference  of  the  two  given  vectors,  in 
direction  and  magnitude,  and  thus  deter- 
mines   the   sine   wave   of   the   resultant 
voltage.     If  it  were  required  to  subtract 
OB  from  OA,  it  would  be  necessary  to 
draw   AC   equal   and   opposite   to    OB. 
The   resultant   is    OC ';  it   is   equal   and 
opposite    to    the   former   resultant    OC. 
This  is  in  accord  with  the  general  alge- 
braic rule  that  A  —  B  =  —  (B  —  A}.    By 
reference  to  Fig.  5  it  will  be  seen  that  in 
the  case   of   addition   the   order   of   the 
vectors  does  not  affect  the  value  or  the 

direction  of  the  resultant.    This  is  also  according  to  the  algebraic 
rule  that  A  +  B  =  B  +  A. 


FIG.  6.   Subtraction  of 
vectors. 


24  SINE  WA  VES  AND  VECTORS.  [CHAP.  2. 

The  preceding  results  with  regard  to  the  addition  and  subtrac- 
tion of  vectors  can  be  summed  up  in  the  following  rule  :  Relations 
which  are  true  algebraically  for  instantaneous  values  of  sinusoidal 
currents  and  voltages  hold  true  geometrically  for  the  vectors  of 
these  quantities.  It  is  customary  to  provide  vectors  of  currents 
with  triangular  arrows,  as  in  Fig.  5  ;  vectors  of  voltages  are 
usually  distinguished  by  pointed  arrows,  as  in  Fig.  6.  This 
distinction  enables  one  to  see  directly  from  the  diagram  whether 
a  vector  represents  a  current  or  a  voltage,  without  reference  to 
the  text. 

PROB.  32.  Construct  the  vector  of  the  resultant  current  re- 
quired in  problems  29  and  30,  and  check  its  magnitude  and  phase 
position  with  those  obtained  analytically. 

PROB.  33.  Two  alternators,  with  the  same  number  of  poles, 
are  coupled  together  so  as  to  give  voltages  differing  in  phase  by 
a  =  27  degrees,  the  voltage  of  the  second  machine  leading  that 
of  the  first  by  this  angle.  The  first  alternator  generates  voltage 
Em  =  2300  volt.,  the  second  Em'  =  1800  volts.  The  two  machines 
are  connected  electrically  in  series.  Find  graphically  the  vector 
of  the  resultant  voltage  in  its  magnitude  and  phase  position. 
Find  also  the  vector  of  the  resultant  voltage  when  the  terminals 
of  one  of  the  machines  are  reversed.  Ans.  (i)  3988  v.,  leading 
Em  by  n°  49';  (2)  1074  v.,  lagging  behind  Em  by  49°  32'. 

PROB.  34.  An  alternator,  the  terminal  voltage  of  which  is 
^=6600  volt.,  supplies  a  load  through  a  transmission  line. 
The  conditions  are  such  that  the  current  lags  behind  the  genera- 
tor voltage  by  an  angle  <j>  =  35  degrees.  The  voltage  drop  in 
the  line  is  e  =  540  volt.,  leading  the  current  in  phase  by  an 
angle  a  =  67  degrees.  Find  the  receiver  voltage  E' ',  by  subtract- 
ing the  voltage  drop  in  the  line  from  the  generator  voltage 
(geometrically);  also  determine  the  phase  displacement  <£'  be- 
tween the  receiver  voltage  and  the  current.  Ans.  E'  =  6149 
volt.  <£'  =  32°  20'. 


CHAPTER  III. 

POWER  IN  ALTERNATING-CURRENT  CIRCUITS. 

9.  Power  when  Current  and  Voltage  are  in  Phase.  Let 
a  resistance  r  be  connected  across  the  terminals  of  an  alternator, 
the  voltage  at  the  terminals  varying  according  to  the  sine  law. 
The  current  through  the  resistance  also  varies  according  to  the 
sine  law,  because  Ohm's  law  holds  true  for  any  moment  of  time, 
so  that  the  curve  of  current  is  in  phase  with  that  of  the  voltage. 
If  the  equation  of  the  voltage  wave  is  e  =  Em  sin  x,  the  equation 
of  the  current  is  i  =  EJr  •  sin  x.  Graphically,  the  current  and 
the  voltage  are  represented  by  two  vectors  of  different  length, 
but  in  the  same  direction  ;  for  instance,  like  OCand  OD  in  Fig.  6. 

Divide  the  time  T  of  one  cycle  into  a  large  number  of  small 
intervals  A/.  Then  the  amount  of  energy  delivered  to  the  re- 
sistance r  and  converted  into  Joulean  heat  during  one  of  such 
intervals  varies  with  the  time- position  of  the  interval  in  the 
cycle  ;  in  other  words,  with  the  instantaneous  values  of  the  volt- 
age and  the  current.  This  energy  is  practically  equal  to  zero, 
when  the  current  and  the  voltage  have  values  near  zero,  and  it 
reaches  a  maximum  with  them.  However,  the  energy  never 
becomes  negative,  because  whether  the  current  flows  in  one 
direction,  or  in  the  opposite,  the  heat  liberated,  z'V-Atf,  is  always 
positive. 

It  is  customary  to  speak  of  the  rate  of  liberation  of  energy 
per  unit  time>  instead  of  the  actual  amounts  of  energy  during 
very  short  or  infinitesimal  intervals  of  time.  It  must  be  under- 
stood, of  course,  that  with  alternating  currents  the  rate  is 
variable,  and,  strictly  speaking,  applies  only  to  one  moment  of 
time.  Let  the  instantaneous  value  of  the  current  at  a  certain 
instant  t  be  i.  If  this  current  remained  constant  for  one  second, 
the  energy  liberated  would  be  equal  to  z  V.  As  a  matter  of  fact, 
this  current  can  be  considered  constant  only  during  the  infini- 
tesimal element  of  time  dty  so  that  the  energy  liberated  during 
this  element  is  i*r*dt.  Nevertheless,  it  is  proper  to  say  that  at 
the  instant  under  consideration  the  energy  is  liberated  at  the  rate 
equal  to  z'V  per  unit  time,  because  z'V«  *///*#=  z  V.  This  is  ana- 


26  POWER  IN  A.  C  CIRCUITS.  [CHAP.  3. 

logous  to  speaking  of  the  instantaneous  speed  of  a  body  during 
a  period  of  acceleration  or  retardation.  The  speed  varies  from 
instant  to  instant,  so  that  to  say  that  the  speed  is  v  at  a  certain 
instant  merely  means  that,  z/"the  body  continued  to  move  at  this 
velocity  for  one  second,  it  would  cover  a  space  equal  to  v. 

The  rate  at  which  energy  is  liberated,  or  the  energy  per  unit 
time,  is  called  power.  Thus,  it  is  permissible  to  speak  of  instan- 
taneous power  in  the  sense  in  which  we  speak  of  instantaneous 
values  of  speed  in  a  non-uniform  motion.  The  instantaneous 
power  indicates  the  amount  of  energy  which  would  be  developed 
per  second,  z/"the  current  suddenly  became  constant. 

The  total  energy  liberated  in  the  form  of  heat  during  one  com- 
plete cycle  is 


W 


/T  (*T 

i*r  •  dt  =  r  I     i z  •  dt.      .      .      .     (24) 
o  J  o 


Another  expression  for  power  in  a  direct-current  circuit  is  ei. 
In  an  alternating-current  circuit,  the  voltage  and  the  current 
can  be  assumed  constant  during  an  infinitesimal  interval  of  time 
dt.  Therefore,  the  instantaneous  energy  is  ei>dt,  and  the  energy 
developed  during  one  cycle  is 

ei  •  dt.      .     .     .     .      .      (24a) 


=    f 

J 


PROB.  35.  The  current  given  in  problem  16  flows  through  a 
resistance  of  10  ohms.  Plot  curves  of  instantaneous  values  of 
voltage  and  power.  Ans.  Em  =  750  volt.  ;  max.  power  =  56.  25  kw. 

PROB.  36.  Determine  total  energy  liberated  per  cycle  in  the 
preceding  problem  by  integrating  graphically  the  curve  of  power. 
Ans.  562.5  joules  (watt-seconds). 

PROB.  37.  Prove  analytically  that  the  curve  of  power,  ob- 
tained in  problem  35,  is  a  sine  wave  of  double  frequency,  tangent 
to  the  axis  of  time.  Proof  :  The  equation  of  the  curve  is 
P  =  7mV  •  sin  2x.  But  from  trigonometry 

cos  2x  =  cos  *x  —  sin  *x  =  i  —  2  sin  *x. 

Substituting  this  value  of  sin  *x  into  the  expression  for  P  we 
get: 


CHAP.  3.]  POWER  IN  A.  C.  CIRCUITS.  27 

The  first  term  is  constant,  the  second  represents  a  sine  wave  of 
double  frequency,  because  2x  =  2?r  (2/)  /.  The  first  term  is 
always  larger  than  the  second,  so  that  P  is  always  positive,  and 
the  whole  curve  lies  above  the  axis  of  abscissae.  Only  when  zx 
is  a  multiple  of  2-rr  the  second  term  becomes  equal  to  the  first  and 
P=  o.  At  these  points  the  curve  is  tangent  to  the  axis  of 
abscissae. 

PROB.  38.   Integrate  expression  (24)  for  the  energy  per  cycle 
when  the  current  is  expressed  by  eq.  (18).     Solution  : 

JT  ^r 

f'h&»/msr  I     sin2*  •  dt. 
o  */   o 

But  x  —  2-irtlT]  consequently 

dt=  7)27r  •  dx.  -  .     .    .,     .     .     .     (25) 

When  t  =  o,  x  —  o  ;  when  /  =  T,  x  =  2ir.  We  have  therefore, 
changing  the  limits  of  integration  : 


T*        ri          C2*  •     2 
=  72m  •  7/27T  •  I       sin  2; 

•/    o 


dx. 


This  integral  is  best  calculated  by  observing  that  its  value  re- 
mains the  same  if  cosine  is  substituted  for  the  sine.  This  is 
because  integration  is  a  summation,  and  in  summing  up  sines  or 
cosines  over  27r  one  goes  through  the  same  values,  only  in  a 
different  order.  Thus,  we  have  : 

/2TT  x»27T 

sin  *x  -  dx  =  Q ;  cos  *x  •  dx  =  £>, 

o  J  o 

where  Q  is  the  unknown  value  of  both  integrals.  Adding  these 
integrals  together,  and  remembering  that  sin  *x  -f  cos  *x  —  i ,  we 
obtain  : 

,2ir 

dx  =  2?r  =  2Q, 


f 

J  o 


or,  Q  —  TT.     Hence, 


21T 

sin  2x  •  dx  —  IT (26) 


O 

Consequently, 


T. (27) 


28  POWER  IN  A.  C.  CIRCUITS.  [CHAP.  3. 

PROB.  39.  Perform  the  integration  required  in  the  preceding 
problem  by  replacing  sin  *x  by  the  cosine  of  the  double  angle  ; 
see  problem  37. 

In  practice,  the  average  rate  at  which  energy  is  delivered  is  of 
interest,  in  other  words,  the  average  value  of  the  variable  power. 
This  is  analogous  to  using  in  calculations  the  average  speed  of  a 
machine,  when  the  actual  speed  varies  within  certain  limits. 
This  average  power  is  found  by  dividing  the  total  energy 
developed  during  one  cycle  by  the  time  T  of  one  cycle.  When 
the  current  varies  according  to  the  sine  law  the  total  energy  per 
cycle  is  expressed  by  eq.  (27).  Dividing  both  sides  by  T  we 
find  that  the  average  power 

P=\T'mr  ........    (28) 

Had  we  started  with  eq.  (24a)  instead  of  eq.  (24)  we  would 
have  arrived  at  the  result 


n-     ......     (28a) 

Expressions  (28)  and  (28a)  are  identical,  because  Em  =  r/m, 
according  to  Ohm's  law.  It  is  convenient  to  use  in  eq.  (28)  a 
new  value  of  the  current,  /=  /m/v/2,  instead  of  7m,  because  then 
the  expression  for  power  becomes  identical  with  that  used  with 
direct  currents,  namely 

P=Pr  ........      (29) 

A  constant  current  /  so  chosen  that  its  square  is  equal  to  the 
mean  value  of  the  squares  of  instantaneous  values  of  a  given 
alternating  current,  is  called  the  effective  value  of  the  alternating 
current.  It  is  seen  from  formulae  (28)  and  (29)  that  for  sinu- 
soidal currents  the  effective  value  is  equal  to  the  amplitude 
divided  by  the  square  root  of  two,  or 


The  effective  value,  as  defined  above,  gives  the  same  average 
power  and  the  same  total  energy  during  one  cycle  as  the  actual 
alternating  current  which  it  represents.  We  have 

7V-  T=r  £Ti*  -  dt 


CHAP.  3.]  POWER  IN  A.  C.  CIRCUITS.  29 

or 

/»=    CTz*dtlT. (3oa) 

•f  o 

The  last  equation  expresses  in  mathematical  language  that  P 
is  the  average  value  of  i2  over  a  cycle.  Taking  the  square  root 
of  both  sides  of  this  equation,  we  can  also  define  the  effective 
value  /  as  the  square  root  of  the  mean  square  of  instantaneous 
values.  This  definition  is  true  for  any  form  of  alternating  waves.* 

PROB.  40.  Figure  out  the  effective  value  of  the  current  in 
problem  35,  and  show  that  the  expression  Pr  is  equal  to  the 
average  ordinate  of  the  power  curve,  plotted  before. 

Power  is  also  expressed  in  direct-current  circuits  by  <?2/r. 
Analogously,  total  energy  per  cycle  of  an  alternating-current 
circuit  can  be  expressed  by 


W 


/T 
e*\r  •  dt, 
Q 


where  e  is  the  instantaneous  value  of  the  voltage.  It  will  be 
remembered,  however,  that  this  discussion  is  limited  to  the  case 
where  the  circuit  consists  only  of  resistance  r,  and  contains  no 
counter-electromotive  forces,  caused  by  inductance,  capacity,  or 
motors.  Only  under  this  assumption  the  relation  i  —  e\r  holds 
true  for  each  instant,  and  consequently  e*lr  represents  the  rate 
of  dissipation  of  energy,  (see  the  end  of  Sec.  i). 

Here  again,  it  is  convenient  to  introduce  the  effective  value  E 
of  the  voltage,  such  that  the  total  energy  per  cycle  could  be 
represented  by  E2lr  •  T\  in  other  words,  E*\r  should  give  the 
average  power.  Since  r  is  constant,  we  see  that  the  effective 
value  of  the  voltage  is  defined  in  exactly  the  same  manner  as  the 
effective  value  of  the  current,  namely,  that  its  square  is  equal  to 
the  mean  value  of  the  squares  of  instantaneous  values  e.  Hence, 
for  sinusoidal  voltages 

E=EJ^2.     .     .     ;    .     ....     (31) 

Power  is  also  expressed  in  direct-current  circuits  by  e.i.  so 
that  it  is  natural  to  expect  that  in  alternating-current  circuits, 
without  phase  displacement,  the  average  power  should  be  equal 


*This  operation  can  be  performed  graphically  in  rectangular  co-ordinates 
by  plotting  a  curve  of  i2,  or  in  polar  co-ordinates  by  plotting  a  curve  of  * 
directly. 


30  POWER  IN  A.  C.  CIRCUITS.  [CHAP.  3. 

to  the  product  E  •  /of  the  effective  values.  This  is  easily  veri- 
fied by  substituting  the  values  of  E  and  /  from  eqs.  (30)  and 
(31)  into  eq.  (27).  We  have 

W=IjV2-Inrl</2-T=I-E.T  .  .  .  (32) 
To  find  the  average  power  divide  both  sides  of  this  equation  by 
T ;  this  gives 

/»->•/.     .     .     .     ...     .     .     (33) 

which  was  to  be  proved. 

PROB.  41.  Prove  that  the  average  power  with  sinusoidal  cur- 
rents and  voltages  is  equal  to  \ImEm,  or  equal  to  E  -  /,  by  directly 
integrating  the  expression  ei  •  dt  for  infinitesimal  energy.  Use 
the  method  shown  in  problem  38. 

The  preceding  discussion  and  examples  show  the  advantages 
of  using  the  effective  values  of  currents  and  voltages,  instead 
of  the  amplitudes,  when  drawing  vectors.  The  chief  advant- 
age is  that  the  average  power  can  be  calculated  as  in  direct-cur- 
rent circuits,  without  dividing  the  product  of  the  amplitudes  by 
some  factor,  for  instance  2  in  case  of  the  quantities  varying 
according  to  the  sine  law.  Commercial  ammeters  and  voltmeters, 
intended  for  use  on  alternating- current  circuits,  are  always 
calibrated  to  indicate  the  effective  values  of  currents  and  voltages 
(See  the  author's  Experimental  Electrical  Engineering,  pp.  38, 
44,  and  47). 

PROB.  42.  An  electric  heater  was  tested  for  power  consump- 
tion on  an  alternating-current  circuit,  by  having  an  ammeter  in 
series  with  it,  and  a  voltmeter  across  its  terminals.  Both  instru- 
ments were  calibrated  to  indicate  effective  values.  The  readings 
were  :  E—  no  volts.,  /=  5.7  amp.  Assuming  the  current  and 
the  voltage  to  have  been  in  phase,  which  is  nearly  the  case,  what 
was  the  average  power  consumption  of  the  heater,  and  what  was 
its  resistance?  Determine  also  the  maximum  instantaneous 
values  of  the  current  and  the  voltage,  under  the  supposition  of 
the  sine  law.  Ans.  P=  627  watts  ;  r  =  19.3  ohm.;  Em—  155.56 
volt.;  7m  =  8.06  amp. 

10.  Power  when  Current  and  Voltage  are  out  of  Phase. 
In  most  practical  alternating-current  circuits  there  is  a  more  or 
less  pronounced  phase  displacement  between  the  current  and  the 
voltage.  This  is  due  to  the  presence  of  counter-electromotive 


CHAP.  3.]  POWER  IN  A.   C.  CIRCUITS.  31 

forces  in  the  circuit,  the  principal  among  these  being  :  (a)  The 
counter-electromotive  forces  of  the  motors  connected  into  the 
circuit,  (b)  The  counter-electromotive  forces  induced  by  alter- 
nating magnetic  fluxes  in  the  circuit.  These  fluxes  may  be 
those  created  by  the  current  itself,  or  they  may  be  due  to  the 
influence  of  other  circuits  (self-  and  mutual  induction),  (c) 
Counter-electromotive  forces  due  to  the  elasticity  of  the  dielec- 
tric medium  surrounding  the  circuit  (electrostatic  capacity). 

The  actual  workings  of  these  causes  are  discussed  more  in 
detail  in  the  following  chapters  and  in  the  comparsion  pamphlet 
on  "The  Magnetic  Circuit."  Here  it  is  sufficient  to  note 
that  there  are  factors  producing  counter-electromotive  forces 
in  alternating-current  circuits,  and  that  they  bring  about  a 
phase  displacement  between  the  voltage  and  the  current.  The 
cause  of  this  displacement  can  be  understood  with  reference  to 
Fig.  6.  Let  OB  be  the  generator  voltage,  and  let  OA  represent 
that  part  of  OB  which  is  equal  and  opposite  to  the  sum  of  the 
various  counter-electromotive  forces  in  the  circuit.  Subtracting 
OA  from  OB  the  net  voltage  OC  is  obtained  which  is  just  suffi- 
cient to  supply  the  ohmic  drop  in  the  circuit.  The  current  OD 
is  in  phase  with  this  voltage,  and  is  numerically  equal  to  OC 
divided  by  the  total  resistance  r  of  the  circuit.  It  will  be  seen 
that  there  is  a  phase  displacement  <f>  between  the  current  and  the 
generator  voltage  OB  ;  it  is  clear  from  the  figure  that  this  phase 
displacement  is  due  to  the  presence  of  the  counter-electromotive 
force  BC. 

Consider  first  the  specific  case  when  the  phase  displacement 
between  the  current  and  the  voltage  is  exactly  90  degrees.  If 
the  current  is  represented  by  eq.  (18),  the  expression  for  the 
voltage  is  e  =  E^  cos  x.  The  instantaneous  power  is  equal  to 
i  •  c  —  IJE^  sin  x  cos  x  =  IJSJ2  •  sin  2x.  Thus,  the  power  varies 
as  a  sine  function  of  double  the  generator  frequency  ;  the  energy 
flows  now  away,  and  now  toward  the  generator.  The  average 
power  for  one  cycle  is  therefore  zero,  for  the  power  has  all  the 
negative  values  that  it  has  positive  ones.  Mathematically,  this 
result  is  represented  by  the  time  integral  of  the  instantaneous 
power  over  a  complete  cycle.  Omitting  the  constant  quantities 
Em  and  7m  we  have 

j      sin  .r  cos  *  d^r  =  £     — •  cos  2#        =  o.     .     .       (34) 


32  POWER  IN  A.  C.  CIRCUITS.  [CHAP.  3. 

L,et  now  the  phase  displacement  between  the  current  and  the 
voltage  be  less  than  90  degrees,  and  be  equal,  say,  to  <f>.  The 
average  power  delivered  by  the  alternator  is  in  this  case  smaller 
than  the  product  El,  and  its  value  must  be  deduced  separately. 
The  voltage  Em  can  be  resolved  into  a  component  Em  cos  <£  in 
phase  with  the  current,  and  another  component,  Em  sin  <£,  in 
quadrature  with  the  current.  According  to  the  proof  given 
above,  the  power  .produced  by  the  second  component  of  the 
voltage  is  zero,  so  that  the  total  average  power  is  [see  eq.  (28a)]  : 
/>=^mcos*./m  =  i^m/ncos*.  .  .  .  (35) 

Replacing  the  amplitudes  by  the  effective  values,  this  equation 
is  converted  into  the  usual  expression  for  the  average  power  : 

P=  El-  cos<£.     ......      (36) 

A  more  rigid  deduction  of  this  expression  is  given  in  problem  45 
below. 

The  product  El  is  sometimes  called  the  apparent  power,  and 
cos  <f>  is  usually  referred  to  as  the  power  factor.  Thus,  the  power 
factor  can  be  denned  either  as  the  cosine  of  the  angle  of  phase 
displacement  between  the  current  and  the  voltage,  or  as  the  ratio 
of  the  true  power  P  to  the  apparent  power  IE. 

Referring  to  Fig.  6,  the  factor  /  cos  <f>  which  enters  into  eq. 
(36)  represents  the  projection  of  /  upon  the  direction  of  the 
voltage  OB,  or  E.  Hence,  eq.  (36)  can  be  interpreted  by  saying 
that  the  true  power  is  equal  to  the  product  of  the  voltage  by  the 
component  of  the  current  in  phase  with  it.  This  component  of 
the  current  is  called  therefore  the  energy  component,  while  the 
component  of  the  current  at  right  angles,  or  in  quadrature  with 
the  voltage,  is  called  the  wattless  component.  The  quadrature 
component  is  wattless  because  the  cosine  of  the  angle  between  it 
and  the  voltage  is  zero. 

Instead  of  resolving  the  vector  of  the  current  into  components, 
it  is  sometimes  preferable  to  resolve  the  voltage  E  into  the  com- 
ponents E  cos  <£  and  E  sin  <£,  in  phase  and  in  quadrature  with 
the  current.  In  this  case,  eq.  (36)  is  expressed  in  words  by 
saying  that  the  average  power  is  equal  to  the  current  times  the 
component  of  the  voltage  in  phase  with  it.  These  components 
of  the  voltage  are  called  the  energy  component  and  the  wattless 
component  of  the  voltage  respectively. 


CHAP.  3.]  POWER  IN  A.  C.  CIRCUITS.  33 

PROB.  43.  Assuming  the  line  current  in  problem  34  to  be  452 
effective  amperes,  calculate  the  average  power  Pl  delivered  by 
the  alternator,  and  the  power  P2  received  at  the  opposite  end  of 
the  line.  P1  =  2444  kw.;  P2  =  2350  kw. 

PROB.  44.  Referring  to  problem  42,  a  wattmeter  was  connected 
into  the  heater  circuit,  and  the  true  power  was  measured  to  be 
598  watts.  Assuming  all  the  three  instruments  to  be  in  calibra- 
tion, calculate  the  power  factor  and  the  angle  of  displacement 
between  the  current  and  the  voltage  in  the  heater ;  also  the 
energy  and  the  wattless  components  of  the  current.  Ans.  cos  <f>  = 
95.38  per  cent ;  <£  =  17°  30';  5. 388  amp.;  1.714  amp. 

PROB.  45.  Deduce  expression  (35)  for  power  in  inductive 
circuits  directly  from  the  general  expression  (24a).  Solution  : 
Let  the  current  be  expressed  by  eq.  (18),  and  let  the  voltage  be 
leading  by  an  angle  <J>.  The  expression  for  the  voltage  is  then 

Substituting  these  values  into  eq.  (243)  we  get 

I     eidt  —  E^Im  I     sin  x  sin  (x  +  <£)  dt 
J  o  '    •/  o 

=  El  T/2TT  •  I       sin  x\  sin  x  cos  <£  -f  cos  x  sin  <£  \dx 

Jo  J 

=  EJn  7727r  •    cos  <H      sin 2*  dx 

*J   o 

+  sin<£J      sin  x  cos  xdx\. 

The  value  of  the  first  integral  is  IT  [see  eq.  (26)]  ;  the  second 
integral  is  equal  to  zero,  according  to  eq.  (34).  Substituting 
these  values  and  dividing  both  sides  of  the  equation  by  T, 
formula  (35)  is  obtained. 

PROB.  46.  Draw  sine  waves  of  the  current  and  of  the  voltage 
in  problem  27  ;  also  plot  the  curve  of  instantaneous  values  of 
power.  Check  the  average  ordinate  of  the  power  curve  with 
formula  (35).  Explain  how  power  is  negative  during  a  part  of 
the  cycle,  remembering  that  there  are  counter-electromotive 
forces  in  the  circuit. 


34  POWER  IN  A.  C.  CIRCUITS.  [CHAP.  3. 

PROB.  47.  Prove  that  .  the  curve  of  power  consists  of  a  sine 
wave  of  double  frequency,  plus  a  constant  term,  the  latter  repre- 
senting the  average  power.  Suggestion  :  Compare  with  problem 
37,  ie  =  fm£m  sin  x  sin  (x  +  <£).  Use  the  trigonometric  trans- 
formation :  2  sin  A  sin  B  —  cos  (A  —  B}  —  cos  (A  4  B). 

PROB.  48.  Determine  the  ratio  of  the  average  ordinate  of  a 
sine  curve  to  its  amplitude.  Solution  :  The  average  ordinate 
for  one  quarter  of  the  cycle  is  determined  from  the  condition 


whence 

nve/^m=2/T=    '6366.         .         .         .         .         (38) 

PROB.  49.  The  ratio  of  the  effective  value  of  an  alternating 
current  or  voltage  to  its  average  value  is  called  the  form  factor. 
Determine  the  values  of  the  form  factor  for  the  rectangular 
wave,  the  sine  wave,  and  the  triangular  wave.  Ans.  i.ooo; 
i.  in  ;  1.155. 


CHAPTER  IV. 

REACTANCE    AND    RESISTANCE   IN   ALTERNATING- 
CURRENT  CIRCUITS. 

ii.  Inductance  and  Reactance.  This  chapter  is  intended 
to  supplement  Chapter  V  of  the  author's  "  Experimental  Elec- 
trical Engineering."  The  student  is  referred  to  that  book  for 
the  physical  concepts  and  definitions  of  inductance  and  reac- 
tance, and  for  the  fundamental  experiments  which  illustrate  the 
properties  of  conductors  in  this  respect.  As  a  matter  of  fact,  it 
is  not  necessary  for  the  understanding  of  this  chapter  to  know 
that  the  cause  of  the  counter-electromotive  force  of  inductance 
is  the  magnetic  flux.  It  is  sufficient  to  accept  as  an  ex- 
perimental fact  that  electricity  in  motion  possesses  some  kind 
of  inertia.  This  inertia  is  analogous  to  that  of  a  fluid  in  a 
pipe.  When  the  motion  of  the  fluid  is  retarded  or  accelerated 
by  an  external  force,  the  inertia  brings  into  play  a  reactive  force 
which  tends  to  oppose  the  change  in  velocity. 

In  the  case  of  an  alternating  current,  which  varies  according 
to  the  sine  law,  this  reaction  or  counter-electromotive  force  also 
varies  according  to  the  sine  law.  In  practice,  it  is  often  more 
convenient  to  consider  the  voltage  applied  at  the  terminals  of  the 
reactance,  equal  and  opposite  to  the  above  mentioned  counter- 
electromotive  force.  This  voltage  is  expressed  by  equation  (3) 
on  p.  118  of  the  "  Experimental  Electrical  Engineering."  This1 
equation,  as  well  as  eq.  (4)  containing  the  definition  of  reactance, 
form  the  basis  of  all  the  further  discussion ;  it  is  therefore 
essential  that  the  student  should  make  their  significance  clear  to 
himself,  before  proceeding  further.  The  fundamental  relations 
are  also  explained  in  Figs.  102,  108,  and  109.  The  results 
reached  are  briefly  as  follows  : 

(i)  When  an  alternating  current  is  flowing  through  a  con- 
ductor, the  alternating  flux  created  around  the  conductor  induces 
in  it  a  counter-electromotive  force.  This  electromotive  force 
lags  in  phase  by  90  degrees  with  respect  to  the  current,  and  is 
numerically  equal  to  2-nfLI.  Here  /  is  the  effective  value  of  the 
current,  in  amperes ;  the  voltage  is  expressed  in  effective  volts. 


36  REACTANCE  AND  RESISTANCE.  [CHAP.  4. 

L  is  a  coefficient  which  characterizes  the  conductor  with  respect 
to  its  ability  to  create  counter- electromotive  force,  with  changing 
current.*  In  this  chapter,  the  inductance  L  is  supposed  to  be 
given  or  measured  experimentally. 

(2)  The  voltage  applied  at  the  terminals  of  the  inductance  is 
equal  and  opposite  to  the  counter-electromotive  force,  and  con- 
sequently leads  the  current  by  90  degrees. 

(3)  The  quantity  zir/L  which  enters  into  voltage  calculations 
is  called  the  reactance  of  a  conductor  or  a  winding.    The  reactance 
multiplied  by  the  current  /  gives  the  voltage  drop  in  the  con- 
ductor ;    therefore,    the   reactance   is   measured    in    ohms, 'like 
resistance. 

(4)  When  a  conductor  possesses  both  resistance  and  reactance, 
or  when  a  resistance  and  a  reactance  are  connected  in  series,  the 
total  voltage  drop  is  represented  by  the  hypothenuse  of  a  right 
triangle,  the  two  other  sides  representing  the  ohmic  and   the 
inductive  drops  respectively.     Consequently,  resistance  and  re- 
actance must  be  added  at  right  angles,  as  is  shown  in  Fig.    108  ; 
their  combined  effect  is  called  the  impedance  of  the  conductor. 

(5)  Total  voltage  drop  across  a  conductor,  which  possesses 
both  resistance  and  reactance,  is  equal  to  the  impedance  of  the 
conductor  multiplied  by  the  current.     Referring  to  Fig.  109,  it 
will  be  seen  that  the  power  factor,  cos  <£,  is  equal  to  ir\iz  —  r\z. 
In  other  words,  the  ratio  of  the  resistance  to  the  impedance  gives 
the  power  factor,  cos  <£,   and  thus  determines  the  angle  <£  by 
which  the  current  lags  behind  the  voltage. 

PROB.  50.  The  inductance  of  a  coil  is  .2  henry,  its  ohmic  re- 
sistance is  negligible.  Draw  a  curve  giving  the  voltage  necessary 
to  maintain  a  current  of  12  amps,  through  the  coil,  at  frequencies 
ranging  from  zero  to  100  cycles  per  second.  Ans.  A  straight 
line  through  the  origin  ;  at/=  100,  E  =  1508  volts. 

P;ROB.  51.  A  reactive  coil  without  iron  draws  a  current  of  75 
amps,  when  connected  across  a  no  volt,  25  cycle  circuit.  What 
current  would  it  draw  at  60  cycles  and  at  the  same  voltage,  pro- 
vided that  the  effect  of  its  resistance  can  be  neglected  ?  Plot  a 
curve  of  current  at  intermediate  frequencies.  Ans.  31.25  amps.; 
equilateral  hyperbola  asymptotic  to  both  axis. 

*  The  physical  meaning  of  L  in  terms  of  magnetic  lines  of  force,  and  its 
calculation  are  treated  in  the  pamphlet  on  "The  Magnetic  Circuit." 


CHAP.  4.]  REACTANCE  AND  RESISTANCE.  37 

PROB.  52.  A  2200  v.,  600  kw,  50  cycle  transformer  must  have 
a  wattless  magnetizing  current  of  not  over  2.5  per  cent  of  the 
full-load  current.  What  is  the  lower  limit  of  its  no-load  reactance 
and  inductance  ?  Ans.  322.5  ohm  ;  1.027  henry. 

12.  Problems  on  Impedance,  and  Impedances  in  Series. 
The  meaning  of  the  impedance,  and  the  numerical  relations  be- 
tween it,  the  voltage,  and  the  current  are  explained  in  Figs.  102, 
108,  and  109  of  the  Experimental  Electrical  Engineering.  These 
figures  also  show  the  angle  <f>  of  displacement  between  the  voltage 
and  the  current.  The  relations  (10)  and  (n)  on  p.  125  are  also 
useful  in  the  solution  of  problems. 

PROB.  53.  The  coil  considered  in  problem  50  is  connected  in 
series  with  a  100  ohm  resistance  ;  it  is  required  to  maintain  a 
current  of  12  amperes  through  the  two,  at  various  frequencies. 
Supplement  the  curve  obtained  in  problem  50  with  curves  of 
voltage  drop  across  the  resistance,  and  the  total  voltage  across 
the  combination.  Plot  also  the  corresponding  values  of  power 
factor.  Determine  the  ordinates  of  the  curves  graphically,  and 
check  a  few  analytically.  Ans.  Er  =•  1200  v.  independent  of  the 
frequency.  At  /=  o,  Ez  =  Er,  and  cos  <£  =  i.  At  /=  100, 
Ez  =  1927  v.,  cos  <£  =  62.25  per  cent.  The  curves  resemble 
those  in  Fig.  104. 

PROB.  54.  What  is  the  value  of  impedance  in  the  preceding 
problem,  at/  =  100?  Ans.  160.6  ohm. 

PROB.  55.  When  a  certain  non-inductive  resistance  is  connected 
across  a  source  of  alternating  voltage,  a  current  i  flows  through 
it.  When  an  inductance,  containing  negligible  resistance,  is 
connected  across  the  same  source  of  voltage,  the  current  is  i  '. 
What  is  the  current  and  the  phase  displacement  when  the  re- 
sistance and  the  inductance  are  connected  in  series  across  the 
same  source  ?  Solution  :  Let  the  unknown  voltage  be  E.  The 
unknown  resistance  is  r  =  Eli  ;  the  unknown  reactance  x  =•  E\i'  . 
When  the  two  are  connected  in  series,  the  impedance 


Consequently,  the  current  is 

E\z  =  ii'l(i*  +  *")*  ;  tan  <£  =  x\r  =  i\i'  . 


38  REACTANCE  AND  RESISTANCE.  [CHAP.  4. 

PROB.  56.  Supplement  the  curves  in  problem  53  by  those  of 
true  and  apparent  power.  Ans.  14.4  kw.  at  all  frequencies; 
23.12  kv-amp.  at/=  100. 

PROB.  57.  The  instantaneous  readings  on  the  instruments  in  a 
power  house  are:  7520  kw.,  66  kv.;  147  amps.  The  power- 
factor  meter  shows  that  the  current  is  lagging  behind  the  volt- 
age. What  are  the  readings  at  the  same  instant  at  the  receiving 
end  of  the  line,  if  its  resistance  is  45  ohms,  and  its  reactance  83 
ohms.  Ans.  6547  kw. ;  53.4kv.  Hint:  Draw  the  vectors  of  the 
generator  voltage  and  current  in  their  true  relative  position. 
Subtract  the  ohmic  drop  in  phase  with  the  current,  and  the  in- 
ductive drop  in  quadrature  with  it.  The  result  gives  the  receiver 
voltage  in  its  true  magnitude  and  phase  position. 

PROB.  58.  Two  impedance  coils  are  connected  in  series  across 
292  alternating  volts.  The  voltages  across  the  coils  are  152  v. 
and  175  v.;  the  current  is  7.3  amps.  Knowing  that  the  resist- 
ance of  the  first  coil  is  10  ohms,  determine  graphically,  as  in 
Fig.  1 14,  the  resistance  of  the  second  coil  ;  also  the  reactances 
and  the  impedances  of  both  coils.  Ans.  r2  =  12.95  ;  zl  =  20.82  ; 
2^  =  23.97,  aH  m  ohms. 

PROB.  59.  In  order  to  determine  the  power  input  into  a  single- 
phase  no  v.  motor,  without  the  use  of  wattmeter,  the  motor  is 
connected  in  series  with  a  non-inductive  resistance  across  a  220 
v.  circuit.  The  resistance  is  adjusted  so  that  the  voltage  across 
the  motor  terminals  is  no  v.  when  the  motor  is  carrying  the 
required  load.  Under  these  conditions  the  voltage  across  the 
resistance  is  found  to  be  127  v.,  and  the  current  through  the 
motor  23  amps.  From  these  data  determine  graphically  the 
power  factor  of  the  motor,  and  calculate  the  power  input  into  it, 
as  is  explained  in  Sec.  109  of  the  Experimental  Electrical  Engi- 
neering. Ans.  72.27  per  cent  ;  1826  watts. 

PROB.  60.  Referring  to  the  preceding  problem,  calculate 
cos  <f>  trigonometrically,  from  the  triangle  of  voltages,  instead  of 
determining  it  graphically. 

13.  Impedances  in  Parallel.  The  problems  that  follow  are 
intended  to  illustrate  the  theory  given  in  Sections  111-113  of 
the  Experimental  Electrical  Engineering. 

PROB.  61.  Let  the  resistance  and  the  inductance  mentioned  in 
problem  55  above  be  connected  in  parallel.  What  is  the  total 


CHAP.  4.]  REACTANCE  AND  RESISTANCE.  39 

current,  and  the  phase  displacement  between  the  current  and  the 
voltage?     Ans.  /=  (z2  +  z'2)^  ;  tan  <£  =  i'\i. 

PROB.  62.  The  load  of  a  single-phase,  6600  v.  generator  is 
estimated  to  consist  of  1200  kw.  of  lamps,  practically  non-induc- 
tive, and  of  800  kw.  of  motors,  working  at  an  average  power 
factor  of  75  per  cent.  What  will  be  the  expected  generator  out- 
put, in  amperes,  and  the  power  factor  ?  Solution  :  The  energy 
component  of  the  motor  current  is  800/6.6  =  121.2  amp.;  the 
wattless  component  is  121.2  tan  <£  =  106.8  amp.  The  lamp  cur- 
rent is  1200/6.6  =  1 8 1. 8  amp.  The  total  energy  component  of  the 
generator  current  is  121.2+181.8=303  amp.  Consequently, 
total  generator  current  is  (303*+  io6.82)^  =  321.3  amp.;  the 
power  factor  is  303/321.3  =  94.3  per  cent. 

PROB.  63.  Check  the  solution  of  the  preceding  problem  graph- 
ically, as  in  Fig.  121. 

PROB.  64.  Let  the  load  given  in  prob.  62  be  distributed  equally 
among  the  phases  of  a  three-phase  system.  What  is  the  total 
current  per  phase?  Ans.  185.5  amp. 

PROB.  65.  Three  impedance  coils,  having  ohmic  resistances  of 
2,  3,  and  4  ohms,  respectively,  and  inductances  of  13,  10,  and 
22  millihenrys,  are  connected  in  parallel  across  a  source  of  220 
v.,  60  cycle  alternating  voltage.  Construct  the  vector  of  the 
resultant  current,  as  in  Fig.  119.  Ans.  no  amp.;  cos  <£  =  .495. 

PROB.  66.  Solve  the  preceding  problem  for  the  frequency  of 
25  cycles  per  second.  Construct  both  diagrams  to  the  same  scale 
so  as  to  see  the  influence  of  frequency. 

PROB.  67.  In  problem  65  let  the  total  current  be  given  in 
magnitude,  but  not  in  its  phase  position  ;  assume  the  inductance 
of  the  third  coil  to  be  unknown.  Show  how  to  determine 
graphically  the  vector  of  the  current  in  the  third  coil,  and  the 
position  of  the  vector  of  the  total  current. 

14.  Equivalent  Resistance  and  Reactance.  Let  a  resist- 
ance r  and  a  reactance  x  be  connected  in  series  across  a  source 
of  alternating  voltage  E.  Let  another  resistance  rp  and  a  re- 
actance xf  be  connected  in  parallel  across  the  same  source.  The 
latter  resistance  and  reactance  can  be  selected  so  as  to  give  the 
same  total  current  and  the  same  phase  displacement  as  the  first 


40  REACTANCE  AND  RESISTANCE.  [CHAP.  4. 

combination.  The  two  combinations  can  be  called  equivalent, 
and  used  one  instead  of  the  other  in  all  calculations  involving  the 
relations  between  currents  and  voltages.  This  provides  a  method 
of  reducing  complicated  circuits  to  simpler  circuits. 

Let  it,  for  instance,  be  required  to  add  two  impedances  in 
parallel.  Each  impedance  can  be  replaced  by  an  equivalent 
combination  of  a  resistance  and  a  reactance  in  parallel  ;  then  the 
two  resistances  and  the  two  reactances  can  be  replaced  by  one 
resistance  and  one  reactance  respectively.  Finally,  this  parallel 
combination  can  be  replaced  by  an  equivalent  combination  of  a  re- 
sistance and  a  reactance  in  series.  The  details  of  calculations 
are  explained  in  problem  76  below. 

The  problem  is  :  knowing  r  and  x  to  find  rp  and  xp.  The  con- 
ditions for  the  two  combinations  to  be  equivalent  are  expressed 
analytically  as  follows :  The  current  through  rp  is  E\r^  in  phase 
with  E ;  the  current  through  x9  is  E\x^  in  quadrature  with  E. 
Consequently,  the  total  current  through  the  equivalent  combina- 
tion is  (E*lr*  +  E*lx*}Km  This  must  be  equal  to  the  current 
E\z  through  the  original  combination  of  r  and  x  in  series.  Hence, 
equating  the  two,  we  have 

i/^p2  +  i/*pf  =  i/*2 (39) 

In  this  equation  z  is  a  known  quantity,  being  equal  to  (r2  -f  x*Y/2. 
The  angle  of  displacement  between  the  total  current  and  the 
voltage  in  the  equivalent  combination  is  determined  from  the  re- 
lation :  tan  <£  =  (Elx^KEjr^)  —  rjxp.  For  the  given  combina- 
tion in  series  we  have  :  tan  <f>  =  x\r.  Since,  by  assumption,  both 
combinations  are  to  produce  the  same  phase  displacement,  we 
have 

rJx^  =  xlr (40) 

Solving  eqs.  (39)  and  (40)  for  the  unknown  quantities  rp  and 
x  ,  and  remembering  that  r3  +  x*  =  z1,  we  obtain 


PROB.  68.  An  impedance  coil  having  a  resistance  equal  to  2 
ohm.  and  a  reactance  of  7.5  ohm.  are  to  be  replaced  by  an  equiv- 
alent resistance  and  reactance  in  parallel.  Find  their  values. 
Ans.  r  =30. 125  ohm.;  x  =  8.035  ohm. 


CHAP.  4.]  REACTANCE  AND  RESISTANCE.  41 

PROB.  69.  Check  the  solution  of  the  preceding  problem  by 
actually  calculating  the  current  and  the  phase  angle  for  the  two 
combinations,  at  a  certain  assumed  voltage. 

PROB.  70.  Show  that  rp  and  xf  are  always  larger  than  r  and  x 
respectively.  Hint  :  Replace  z*  in  eqs.  (41)  by  r*  +  ,r2. 

PROB.  71.  Let  rp  and  x9  be  given  ;  find  the  values  of  r  and  x. 
Solution  :  Solve  eqs.  (41)  for  r  and  x,  using  the  value  of  z2  cal- 
culated from  eq.  (39). 

PROB.  72.  In  adjusting  a  measuring  instrument  a  non-induc- 
tive resistance  of  120  ohm.  was  used  in  parallel  with  a  choke 
coil.  The  impedance  of  the  coil  was  75  ohm.  its  resistance  16 
ohm.  In  the  regular  manufacture  of  the  instrument  it  is  desired 
to  use  a  resistance  and  a  reactance  in  series.  Determine  their 
values,  either  graphically  or  analytically.  Ans.  r  =  37.9  ohm.; 
x  =  44.15  ohm. 

15.  Admittance  and  its  Components — Conductance  and 
Susceptance.  It  is  shown  in  Sec.  2  that  when  resistances  are 
connected  in  parallel  it  is  more  convenient  to  use  their  recipro- 
cals— conductances.  It  is  natural  to  expect  that  the  same  would 
apply  to  the  addition  of  reactances  and  of  impedances.  This 
leads  to  two  new  concepts,  those  of  susceptance ;  a  reciprocal  of 
reactance,  and  of  admittance,  a  reciprocal  of  impedance.  Since 
reactances  and  impedances  are  measured  in  ohms,  susceptances 
and  admittances  are  measured  in  mhos,  like  conductances. 

We  thus  have  the  following  six  quantities : 

resistance  conductance 

reactance  susceptance 

impedance  admittance 

The  first  three  show  how  difficult  it  is  to  force  a  unit  current 
througk  a  conductor ;  the  last  three  are  the  reciprocals,  and 
show  how  easy  it  is  to  force  a  unit  current  through  the  conductor. 
The  first  three  are  measured  in  ohms,  the  latter  three  in  mhos. 
The  first  three  form  a  right-angle  triangle,  it  is  shown  below 
that  the  latter  three  also  form  a  similar  triangle. 

If  the  resistance  of  a  conductor  is  5  ohms,  its  conductance  is 
0.2  mho.  In  the  same  way,  if  the  reactance  of  a  coil  is  equal  to 
5  ohms,  its  susceptance  is  said  to  be  0.2  mho  (provided,  that  its 
ohmic  resistance  is  negligible).  It  is  necessary  to  distinguish 


42  REACTANCE  AND  RESISTANCE.  [CHAP.  4. 

between  the  conductance  and  the  susceptance,  because  in  one 
case  the  current  is  in  phase  with  the  voltage,  in  the  other  case  it 
is  in  quadrature,  lagging  behind  the  voltage.  If  the  impedance 
of  a  coil  is  5  ohm.,  it  being  understood  that  the  coil  has  both 
ohmic  resistance  and  reactance,  the  admittance  of  the  coil  is  0.2 
mho.  This  admittance  cannot,  however,  be  separated  into  the  con- 
ductance and  the  susceptance,  without  an  additional  definition  of  the 
latter,  even  though  the  resistance  and  the  reactance  of  the  coil 
be  known. 

It  is  customary  to  define  the  conductance  and  the  susceptance  of 
an  impedance  coil  as  those  of  its  equivalent  parallel  combination. 
The  reason  for  this  is  that  conductances  and  susceptances  are 
used  when  impedances  are  added  in  parallel.  As  is  explained 
above,  each  impedance  is  in  this  case  replaced  by  an  equivalent 
resistance  rp  and  reactance  ^p  in  parallel.  The  reciprocals  of 
these  are  called  the  conductance  and  the  susceptance  of  the 
original  coil  ;  it  must  be  remembered,  however  that  in  reality 
they  refer  to  the  equivalent  parallel  combination. 

Let  us  assume,  tor  instance,  that  the  above  mentioned  imped- 
ance coil  possesses  a  resistance  of  3  ohm,  and  a  reactance  of  4 
ohm.  According  to  eqs.  (41),  the  equivalent  resistance  is  25/3 
ohm.,  the  equivalent  reactance  is  25/4  ohm.  Therefore,  by  defi- 
nition, the  conductance  of  the  original  coil  is  3/25  =  .12  mho., 
and  its  susceptance  is  4/25  =  .  16  mho.  It  will  be  seen  that  these 
values  are  different  from  the  erroneous  values  of  1/3  and  1/4 
obtained  by  simply  taking  the  reciprocals  of  the  resistance  and 
the  reactance  of  the  coil  itself. 

PROB.  J2a.  An  apparatus  takes  25  amp.  and  2000  watts  at 
1 10  volt.,  and  the  current  is  lagging.  What  is  the  equivalent 
conductance  and  susceptance  of  the  device  ?  What  is  the  resist- 
ance and  reactance  in  series  equivalent  to  this  apparatus  ?  Ans. 
.165  mho.;  .156  mho.;  3.2  ohm.;  3.02  ohm. 

Susceptances  are  added  in  parallel  like  conductances,  as  may 
be  seen  from  the  following  proof,  similar  to  the  proof  of  eq.  (7). 
Let  several  coils  possessing  reactances  x^  .r2,  etc.,  and  no  ohmic 
resistance  whatever,  be  connected  in  parallel  across  a  source  of 
alternating  voltage  E.  The  currents  through  these  coils  are  : 
z\  =  E\XI  ;  i^  =  E\xv  etc.  All  these  currents  are  in  phase  with 
one  another,  being  displaced  by  90  degrees  with  respect  to  the 


CHAP.  4.]  REACTANCE  AND  RESISTANCE.  43 

voltage  E.  An  equivalent  reactance  x^  is  by  definition  such  as 
would  give  the  same  total  current  as  the  combination  of  the 
given  reactances  in  parallel.  We  have  therefore 


,  or  i*eq  =     i*. 

Denoting  the  reciprocals  of  the  reactances,  or  the  susceptances 
of  the  coils,  by  b,  we  have 

*«,=  *!  +  £,  +  etc  .....  •  .     .     (42) 
This  is  analogous  to  eq.  (7).     We  also  have  for  each  branch 

I=bE.     .     .   '.     /    .     .     .     (43) 

This  corresponds  to  eq.  (6),  except  that  the  current  is  under- 
stood to  be  lagging  behind  the  voltage  by  90  degrees. 

PROB.  73.  Three  resistances  of  2,  5,  and  10  ohm.,  and  two 
reactances  of  4  and  2.5  ohm.  are  all  connected  in  parallel  across 
250  alternating  volts.  WThat  is  the  total  current  and  the  com- 
bined power  factor  ?  Solution:  The  equivalent  conductance  is 
1/2  -f  1/5  -f  i/io  =  .8  mho.  The  equivalent  susceptance  is  1/4  -f 
1/2.5  =  .65  mho.  The  energy  component  of  the  current,  accord- 
ing to  eq.  (6),  is  250  X  .8  =  200  amp.  The  wattless  component, 
according  to  eq.  (43),  is  250  X  .65=  162.5  amp.  Hence,  tan  <£ 
=  162.5/200  =  .8125  ;  cos<£=.775;  total  current  =  200^775  = 
258  amp. 

PROB.  74.  Assuming  the  admittance  of  a  coil  to  be  defined  as 
the  reciprocal  of  its  impedance,  express  this  admittance  through 
the  conductance  and  the  susceptance  of  the  coil.  Ans.  Eq.  (39) 
gives  directly 

/=£-2  +  ^2  .......  (44) 

where  y=  \\z  is  the  admittance. 

PROB.  75.  Express  the  conductance  and  the  susceptance  of  a 
coil  through  its  resistance  and  reactance.  Ans.  Eqs.  (41)  gives 

g  =  r\z>         b^xlz*     .....      (45) 
or,  using  the  admittance^  =  ilz, 

g  =  ry*         b  =  xy\     .     .     .     .     .     (46) 

PROB.  76.  Two  impedances  zl  and  zv  consisting  respectively 
of  resistances  r^  and  rv  and  of  reactances  xl  and  xv  are  connected 
in  parallel.  Find  the  resistance  and  the  reactance  of  the  equiva- 


44  REACTANCE  AND  RESISTANCE.  [CHAP.  4. 

lent  impedance  coil.  Solution  :  Replacing  each  coil  by  a  resis- 
tance and  a  reactance  in  parallel,  and  adding  their  conductances 
and  susceptances,  we  have  :  g^  =  rjzf  -f  r*\z*  ;  similarly  deq  = 
x\\z\  ~t"  XJ2*-  The  admittance  of  the  equivalent  coil  is,  accord- 
ing to  eq.  (44),  y^  =  (g\tl  -f  b\Y/2-  Eqs.  (46)  give  then  :  req  = 


PROB.  77.  Extend  the  solution  of  the  preceding  problem  to 
the  case  of  more  than  two  impedances  in  parallel.  Ans.  g^  = 
'Zrlz*  ;  £eq  =  ^xlz1.  The  rest  of  the  solution  is  the  same. 

PROB.  78.  The  admittance  of  a  winding  is  .2  mho  ;  the  current 
through  the  winding  lags  by  34  degrees  with  respect  to  the 
voltage  at  its  terminals.  Determine  the  resistance  and  the  re- 
actance of  the  winding.  Ans.  ^  =  4.145  ohm.;  x—  2.796  ohm. 

PROB.  79.  A  coil,  having  a  resistance  of  a  2.3  ohm.  and  a 
reactance  of  5  ohm,  is  connected  in  parallel  with  another  coil,  for 
which  r  =  3  ohm.  and  x  =  4  ohm.  Calculate  the  resistance  and 
the  reactance  of  the  equivalent  coil.  Ans.  r=i.T>6  ohm.; 
x=  2.255  ohm. 

PROB.  80.  The  coils  given  in  problem  79  are  connected  in 
parallel  across  55  volts.  Calculate  the  total  current,  its  energy 
and  wattless  components,  and  the  power  factor  of  the  combina- 
tion. Ans.  20.85  amp.;  10.776  amp.;  17.88  amp.;  cos  <£=.  5165. 

PROB.  81.  Prove  that  the  ratio  of  r\x  is  equal  to  the  ratio  of 
gib  ;  in  other  words,  that  the  impedance  triangle  is  similar  to  the 
admittance  triangle. 

PROB.  82.  Check  the  value  of  the  total  current  in  problem  65 
by  calculating  the  combined  admittance  of  the  three  coils. 


CHAPTER  V. 

THE  USE  OF  COMPLEX  QUANTITIES. 

1 6.  Addition  and  Subtraction  of  Vectors  in  Projections. 
With  the  explanation  given  in  the  preceding  four  chapters  the 
student  is  enabled  to  handle,  by  means  of  vector  diagrams,  prob- 
lems involving  resistances  and  reactances  in  alternating-current 
circuits.  Several  problems  in  transmission-line  calculations  and 
in  the  theory  of  alternating-current  machinery  are  reduced  to 
such  vector  diagrams  of  electric  circuits.  The  disadvantages  of 
the  graphical  method  are :  ( i )  Results  are  usually  obtained 
which  hold  for  one  specific  case  only  ;  an  analysis  of  the  effect  of 
various  factors  is  often  difficult.  (2)  Some  vectors  may  be 
many  times  smaller  than  others,  for  instance,  the  voltage  drop  in 
a  line,  as  compared  to  the  line  voltage  itself.  Therefore,  the 
diagram  must  be  drawn  to  a  very  large  scale,  or  else  the  results 
are  not  accurate  enough.  In  addition  to  these  drawbacks,  it 
may  be  said  that  some  people  object  to  graphical  methods  in 
general,  as  involving  the  use  of  drawing  instruments  which  may 
not  be  handy. 

On  the  other  hand,  vector  diagrams  are  quite  convenient  in 
some  practical  cases  ;  moreover,  they  are  helpful  for  the  under- 
standing of  general  relations  in  a  circuit,  without  reference  to 
particular  numerical  values.  Again,  in  some  problems,  the  un- 
known vectors  can  be  calculated  from  the  vector  diagram  tri- 
gonometrically,  without  the  necessity  of  actually  drawing  the 
diagram  to  scale. 

It  is  possible  to  treat  vectors  analytically,  using  their  projec- 
tions on  two  axes,  as  in  analytic 
geometry  (Fig.  7).  A  vector, 
such  as  E,  can  be  defined  either 
by  its  magnitude  and  phase  angle 
<£,  or  it  can  be  given  by  its  pro- 
jections e  and  e'  upon  the  axes 
of  coordinates.  If  R  and  <£  are 


0 


A       given,  the  projections  are  calcu- 
lated from  the  expressions  e  = 
FIG.  7.  A  vector  and  its  projections.    £  cos  $  .   e>  =  E  sin    ^       If  the 


46  COMPLEX  QUANTITIES.  [CHAP.  5. 

projections  are  given,  the  vector  itself  is  determined  by  E  = 
O2  +  *")*,  and  tan  <£  =  «?7*.* 

The  fact  that  e  and  <?'  are  components  of  the  vector  E  along 
two  perpendicular  axes  is  expressed  symbolically  thus  : 

E  =  e+je'.     .     .     .     .     .     .       (47) 

Here/  is  a  symbol  which  indicates  that  the  projection  e'  refers 
to  the  vertical  axis.  This  symbol  must  not  have  any  real  value  ; 
for  the  time  being,  it  may  be  considered  merely  as  an  abbrevia- 
tion of  the  words  :  "  along  the  vertical  axis."  The  sign  plus  in 
eq.  (47)  denotes  the  geometric  addition.  The  dot  under  E  sig- 
nifies that  by  E  is  meant  not  only  the  magnitude  of  the  vector, 

but  its  direction  as  well,  the  latter  being  defined  by  the  projec- 
tions. When  the  magnitude  only  is  meant,  the  dot  is  omitted. 

The  foregoing  notation  has  been  introduced  by  Dr.  Chas.  P. 
Steinmetz,  and  is  now  universally  used  in  this  country,  in  the 
treatment  of  alternating  currents.  Much  credit  is  also  due  to 
Dr.  Steinmetz  for  developing  the  analytic  method  of  dealing  with 
alternating  currents  and  voltages  by  means  of  their  projections. 

Addition  and  substraction  of  vectors  in  projections  is  reduced 
simply  to  the  addition  and  subtraction  of  the  projections  :  Ac- 
cording to  Fig.  5,  the  projection  of  an  equivalent  vector  on  any 
axis  is  equal  to  the  sum  of  the  corresponding  projections  of  the 
component  vectors  on  the  same  axis.  Thus,  if  a  current  is 
represented  as  a  vector  by  its  projections  50  +770  amp.,  and 
another  current  by  1004-740  ainp.,  the  vector  sum  of  these 
currents  is  150  +7110  amp.  Or,  the  resultant  of  two  voltages, 
E  =  e  +£and  E  =  <?  +e'  is 


E^  =  £,  +  E,  =  (e,  +  *, 

As  an  illustration,  let  us  solve  problem  33  of  Chapter  II  by 
the  method  of  projections.  Take  the  vector  of  the  voltage  of 
the  first  alternator  in  the  horizontal  direction,  this  being  the 
simplest  assumption.  This  vector  is  therefore  expressed  as 


*  Instead  of  determining  E  from  the  foregoing  expression  it  is  often  more 
convenient  first  to  determine  the  angle  0  from  tan  <£  =  c'\e,  and  then  calcu- 
late E=ejcos  0,  or  £=  ef\  sin  0,  taking  sine  and  cosine  from  tables.  This 
method  is  used  in  the  solution  of  problem  73,  where  the  total  current  is  cal- 
culated from  its  components. 


CHAP.  5.]  COMPLEX  QUANTITIES.  47 

El  =  2300  +70.     The  horizontal  projection  of  the  second  vector 

is  1800  cos  27°  =  1603.8  v.,  and  its  vertical  projection  is  1800  sin 
27°  =  817.2  v.  Both  of  these  projections  are  positive,  because 
the  second  vector  leads  the  first,  and  is  therefore  in  the  first 
quadrant.  Thus,  E^  —  1603.8  +7817.2  v.  The  resultant  volt- 
age, Ee(i  =  £1  +  E2  =  3903.8  +7817.2  volts.  For  some  purposes, 

it  is  sufficient  to  leave  the  answer  in  this  form,  that  is  to  say,  to 
give  only  the  projections  of  the  resultant  voltage.  If,  however, 
the  magnitude  and  the  phase  position  are  required,  they  can  be 
found  as  explained  above :  tan  <£  =  817.2/3903.8  =  .2092  ;  <£  — 
11°  49'  ;  cos  <£  =  .9786  ;  E^  =  3903.8;. 9786  =  3988  volt. 

If,  however,  the  terminals  of  the  second  machine  be  reversed, 
then,  E^  =  El  —  E2  =  696.2  —7817.2.  This  vector  has  a  posi- 
tive horizontal  projection  and  a  negative  vertical  projection. 
Consequently,  the  vector  lies  in  the  fourth  quadrant,  and  lags 
behind  the  reference  vector  by  less  than  90°.  Preceding  as 
above,  we  find  :  <£  =  —  49°  32',  E^  =  1074  volts. 

PROB.  83.  Solve  problem  34  by  the  method  of  projections, 
assuming  the  vector  of  the  voltage  to  be  horizontal. 

PROB.  84.  Check  the  solution  of  problem  30  by  the  method  of 
projections. 

PROB.  85.  Verify  the  answer  to  problem  65  by  the  method  of 
projections. 

17.  Rotation  of  Vectors  by  Ninety  Degrees.  In  problems 
involving  reactance  it  is  necessary  to  turn  the  vector  of  the 
current  by  90°  and  then  multiply  it  by  the  reactance  in  order  to 
determine  the  reactive  drop  in  voltage.  Multiplication  of  the 
vector  of  the  current  by  the  reactance  converts  it  into  a  vector 
of  voltage,  and  thus  merely  changes  its  scale.  But  turning  the 
vector  modifies  the  relative  magnitudes  of  its  projections  ;  it  is, 
therefore,  necessary  to  find  a  relation  between  the  original  and 
the  new  magnitudes  of  the  projections. 

Let  in  the  simplest  case  a  vector  El  be  drawn  along  the  refer- 
ence axis  of  abscissae,  and  let  its  length  be  a.  In  the  symbolic 
notation  it  is  represented  as  El  =  a,  the  other  projection  being 

zero.  After  having  been  turned  by  90°  counter  clock-wise,  the 
vector  is  directed  along  the  positive  axis  of  ordinates,  and  can 


48 


COMPLEX  QUANTITIES. 


[CHAP.  5. 


be  symbolically  represented  as  Et  =ja,  the  horizontal  projection 
being  zero.  Thus,  in  this  particular  case,  a  rotation  by  90  de- 
grees is  equivalent  to  multiplication  by/. 

//  is  convenient  to  define  j  so  that  multiplication  of  a  vector  by  j 
will  mean  a  rotation  by  90  degrees  in  the  positive  direction 
(counter  clock- wise),  while  division  by  j  will  turn  the  vector  by 
90  degrees  in  the  negative  direction  (clock- wise).  In  order  to 
find  a  value  of/  wh'ich  satisfies  these  requirements,  let  the  vector 
£2  be  turned  again  by  90°  counter  clock-wise,  being  now  directed 

along  the  negative  Jf-axis.     Its  expression   is  now  Es  =  —  a. 

On  the  other  hand,  the  same  expression  must  be  obtained  by 
multiplying  E.z  by/.  Therefore,  we  have  —  a  =/2#,  or/2  =  —  i  ; 

consequently, /  =  v '  —  i.     If   the   original   vector   E^  is   to  be 

turned  by  90°  clock-wise,  we  must,  according  to  our  assumption, 
divide  it  by/.  We  have  :  E^  =  a\j,  or,  multiplying  the  numera- 
tor and  the  denominator  by  /,  E^  =ja\fi  =  — ja.  This  checks 
with  the  preceding  results,  because  E^  =  —  E^.  It  will  thus  be 

seen  that  the  value  of  /2  =  —  i  satisfies  the  requirements  set 
above,  when  the  original  vector  is  directed  along  one  of  the 
axes  of  co-ordinates. 

Let  now  the  original  vector  El  (Fig  8)  have  an  arbitrary  di- 
rection in  the  first  quadrant,  or  El  =  a  -\-  jb.  Multiplying  E  by 

/  we  should  get  the  vector  E^ 
of  the  same  magnitude,  but  in 
the  second   quadrant,  and   per- 
pendicular  to ,  E^.      £2   has  a 
vertical  projection  equal  to  the 
horizontal  projection  a  of  the 
original   vector  E^  ;    the   hori- 
zontal projection  of  E^  is  nega- 
tive, and  is  equal  in  its  absolute 
value  to  the  vertical  projection 
FIG.  8.  Relation  between  the  projec-    b  of    the  vector  Er      Thus,  the 
tions  of  two  vectors,  perpendicular    new     vector     is    expressed    as 
to  each  other.  ^  =  _  b  4.  ja.      On  the  other 

hand,  multiplying   E^  by  /  we  have     jEl=ja-*rj"ib=ja — b. 


CHAP.  5.]  COMPLEX  QUANTITIES.  49 

which  is  the  same  as  above.  Therefore,  in  this  case  also  the 
assumption/2  =  —  i  is  correct  and  leads  to  rotation  by  90  de- 
grees. It  is  left  to  the  student  to  verify  the  cases  when  the 
vector  lies  in  some  other  quadrant,  and  when  Ev  is  divided  by/, 
instead  of  being  multiplied  by/. 

PROS.  86.  A  current  of  80  +743  amps,  flows  through  a  resist- 
ance of  2  ohm.  in  series  with  a  reactance  of  3  ohm.  Find  the 
complex*  expression  for  the  voltage  across  the  impedance. 
Solution  :  The  vector  of  the  voltage  consists  of  two  components, 
representing  the  ohmic  and  the  reactive  drop  respectively.  The 
ohmic  drop,  Elt  is  equal  to  2  (80  +743)  =  160  +786  volt.  To 

find  the  inductive  drop,  Ev  the  vector  of   the  current   must   be 

turned  by  90  degrees,  in  other  words,  multiplied  by  j,  and  then 
multiplied  by  x  =  3.  Thus,  E^  =  $j  (80  +743)  =  —  129  +7240 

volt.     Total  voltage  E  =  E^  -f  Ez  =  31  +7326  volts. 

PROB.  87.  Solve  the  preceding  problem  when  the  voltage  is 
given  and  the  current  unknown.  First  Solution  :  Let  the  un- 
known current  be  represented  by  its  projections  as  i  +  //'.  We 
have,  as  in  the  preceding  problem, 

2  (i  +7*')  +  37  (i  +J*')  =  3i  +7326, 
or,  separating  the  terms  with/, 

(21-  3*')  +/ (2*'  +  30  =  3i  +7326.     .     .      (48) 

This  equation  can  be  satisfied  only  if  the  terms  with  and  with- 
out/ are  equal  to  each  other  separately,  because  a  real  quantity 
cannot  be  increased  or  diminished  in  its  magnitude  by  the  addi- 
tion of  an  imaginary  quantity.  Or,  from  a  geometric  point  of 
view,  the  left  hand  side  and  the  right  hand  side  of  eq.  (48) 
represent  a  vector  in  its  projections.  But  two  vectors  are  iden- 
tical only  when  their  corresponding  projections  are  equal.  Thus, 
we  have 

21  —  3zv  =  31  :     21'  4-  3*  =  326. 

Solving  these  equations  for  i  and  i'  we  find  i  =  80,  i'  =  43,  as  is 
given  in  the  preceding  problem.  Second  Solution  :  The  equation 


*Hxpressions  of  the  form  a  -\-jb,  where  7  is  ^/ —  i,  are  called  in  algebra 
complex  quantities. 


50  COMPLEX  QUANTITIES.  [CHAP.  5. 

preceding  (48)  can  be  written  in  the  form  (2  -|-  37)  (/  4  jif)  = 
31  +7326  ;  or,  i  +  ji'  =  (31  +/326)/(2  -f  37).  Considering  here 
j  as  an  ordinary  algebraic  quantity,  we  can  get  rid  of  it  in  the 
denominator  by  multiplying  the  numerator  and  the  denominator 
by  2  —  37.  The  result  is 


i  .   7%v  ='.(3i  +/326)  (2  -  a/)  _  1040   ,    ;-  559 

'      J  9  x  »X  O  I       _/ 

2  13  J3 


or,  i  -\rji'  =  80  4-/43>  as  before. 

PROB.  8ya.  A  voltage  of  28  +/I2°  volts  applied  to  the  termi- 
nals of  a  coil  produces  in  it  a  current  equal  to  4+71.5  amps. 
Determine  the  resistance  and  the  reactance  of  the  coil.  Ans.  r 
=  16  ohm.;  x  —  24  ohm. 

18.  Impedance  and  Admittance  expressed  as  Complex 
Quantities.  The  method  employed  in  the  solution  of  the  two 
preceding  problems  can  now  be  generalized.  Let  it  be  required 
to  find  the  voltage  necessary  to  maintain  a  current  i  +  ji'  through 
a  resistance  r  and  reactance  x  in  series.  The  voltage  necessary 
for  overcoming  the  resistance  is  equal  to  r  (i+ji')\  that  for 
overcoming  the  counter-e.m.f.  of  the  reactance  i$jx(i+ji'). 
Hence,  the  total  voltage  is  £  =  r  (i  +  ji')  -\-jx  (i  +/&*')>  or 

E  =  e  +je'  =  (r  +  jx)  (i  +  jir).     .     .     .    (49) 

It  is  legitimate  to  factor  out  the  expression  (i  +jir),  and  to 
treaty  as  any  other  algebraic  quantity,  because  j  is  now  assigned 
a  definite  value  ^  —  i.  Moreover,  eq.  (49),  as  well  as  the  equa- 
tion, (48),  represents  simply  the  geometric  addition  of  four 
component  vectors,  two  of  them  being  along  the  Jf-axis  and  the 
other  two  along  the  K-axis.  As  long  as  this  interpretation  is 
kept  in  mind,  the  terms  may  be  arranged  in  any  desired  order. 

Bq.  (49)  shows  that,  in  order  to  obtain  the  expression  of  the 
voltage  corresponding  to  a  current  (i+ji')  through  an  imped- 
ance z  =  (r2  +  ^2)^,  the  current  must  be  multiplied  by  the  com- 
plex expression  r  +jx.  It  must  be  kept  in  mind,  that  r  -\-jx 
is  not  a  vector,  but  merely  an  operator  upon  the  vector  of  the 
current.  The  operation  consists  first  in  multiplying  the  vector 
of  the  current  by  r,  then  in  multiplying  the  same  vector  by  x 
and  turning  it  by  90  degrees  in  the  positive  direction,  and  finally, 


CHAP.  5.]  COMPLEX  QUANTITIES.  51 

in  adding  the  two  vectors  geometrically.  All  these  operations 
are  included  in  the  expression  r  -\-jx. 

In  order  to  get  the  projections  e  and  e'  of  E  from  eq.  (49)  the 

terms  on  the  right-hand  side  must  be  actually  multiplied  and  the 
results  represented  in  the  form  of  a  complex  quantity.  We  get 
then 

E  =  e  +  je'  =  (ri  -  xi'}  +J  (rif  +  xi). 

If  the  voltage  and  the  impedance  are  given,  and  it  is  required 
to  find  the  current,  we  get  from  eq.  (49) 


In  order  to  reduce  the  right  hand  side  of  this  equation  to  the 
form  of  a  complex  quantity  we  multiply  the  numerator  and  the 
denominator  by  the  expression  r  —  jx.  This  gives 

™  +  *<?  +  ft'  -  **     (5o) 
*         >       J  *  ' 


Equating  the  real  and  the  imaginary  parts  gives  the  required 
projections  of  the  current  /. 

Instead  of  dividing  the  voltage  by  the  operator  (r  +  jx}  and 
then  eliminating/  from  the  denominator,  it  is  more  convenient 
to  introduce  another  operator  by  which  the  voltage  must  be 
multiplied  in  order  to  obtain  the  current.  It  will  be  seen  from 
eq.  (50)  that  such  an  operator  is 

I         _  r  —  jx  _        r  x 

~~  ~~ 


Remembering  that  r2  +  x*  =  z*,  and  with  reference  to  eqs.  (45), 
we  see  that  the  foregoing  operator  is  simplified  to  (^  —  /£), 
where  g  is  the  conductance  and  b  the  susceptance  of  the  circuit. 
We  have  thus,  analogously  to  eq.  (49), 

I=i+ji'=(g-jb)  (e+je').     .     .     .      (51) 

The  expression  £•  —  jb  may  be  properly  called  the  inverse  opera- 
tor, because  it  is  equal  to  \\(r  +/*).  The  inverse  operator  is 
convenient  to  use  when  the  voltage  is  given,  while  the 
direct  operator  is  employed  when  the  current  is  known  (see  also 
problem  98  below). 


52  COMPLEX  QUANTITIES.  [CHAP.  5. 

Equation  (49)  can  be  said  to  express  that  the  voltage  E  is 
equal  to  the  product  of  the  current  by  the  impedance,  if  the 
operator  (r  -\-  jx)  be  considered  as  the  impedance  of  the  circuit 
in  thejcomplex  notation.  Denote  the  impedance  by  Z,  then 

Z=r+jx  .......         (52) 

Here  capital  Z  is  used  to  indicate  that  it  is  a  complex  quantity, 
as  distinguished  from  the  numerical  value  z  of  the  same  imped- 
ance. The  letter  is  not  provided  with  a  dot,  because  Z  is  not  a 
vector,  but  an  operator. 

In  an  abbreviated  notation,  eq.  (49)  can  be  written  as 

E=IZ.     .......      (53) 

It  must  be  understood  that  in  this  expression  each  letter  stands 
for  a  complex  quantity,  so  that  when  actual  numerical  or  alge- 
braic relations  are  necessary,  the  expression  must  again  be  ex- 
panded into  (49)  and  the  multiplication  of  the  two  complex 
quantities  actually  performed. 

Similarly,  eq.  (51)  can  be  written  in  the  abbreviated  form  as 

.......       (54) 


where  Kis  the  inverse  operator,  or  the  admittance  of  the  circuit. 
In  the  complex  notation 

y-f-jt  .......    (55) 

Compare  eqs.  (53)  and  (54)  with  eqs.  (i)  and  (6)  of  Chapter  I. 
It  will  be  seen  that  the  former  represent  the  generalized  Ohm's 
law  in  application  to  alternating-current  circuits.  Keeping  the 
meaning  of  the  symbols  in  mind,  it  is  possible  to  solve 
alternating-current  problems  in  the  symbolic  notation  almost  as 
easily  as  direct-current  problems.  The  following  problems  are 
intended  to  illustrate  the  method. 

PROB.  88.  Find,  by  means  of  complex  quantities,  the  voltage 
required  in  problem  57.  Solution  :  Assume  the  vector  of  the 
current  to  be  the  reference  vector.  At  the  power  house  cos  <£  = 
7520/66  X  147  =  .775  ;  sin<£  =  .632  ;  The  generator  voltage 
EI  =  66  X  .775  +/66  X  .632  =  51.15  +741.71  kilovolt.  The 

voltage  drop  in  the  line  is  147  (45  +j  83)  =  6615  +j  1220  volt. 


CHAP.  5.]  COMPLEX  QUANTITIES.  53 

Hence,  the  load  voltage  E=  (51.15  —  6.61)  4-7(41.71  —  12.2)  = 
44.54  4-729.51  kilovolt.  The  numerical  value  of  the  load  voltage 
E=  (44.542  +  29.51')^  =  53-43  kilovolt. 

PROB.  89.  Determine  analytically  the  resistence  r2  required  in 
problem  58. 

PROB.  90.  A  voltage  of  180  +775  volt,  produces  a  current  of 
7  +  7  i. 5  amp.  What  is  the  impedance  of  the  circuit?  Ans. 
26.78  +74.97  ohm. 

PROB.  91.  Power  is  transmitted  from  a  single-phase  alternator 
to  a  load  consisting  of  a  resistance  of  1.17  ohm  in  series  with  a 
reactance  of  .67  ohm.  The  generator  voltage  is  2300  v.,  and  the 
impedance  of  the  transmission  line  =  .085  4-  j  .013  ohm.  Deter- 
mine (a)  the  line  current ;  (b)  the  voltage  drop  in  the  line  ;  (c) 
the  receiver  voltage.  Take  the  generator  voltage  as  the  reference 
vector.  Ans.  (a)  1413.6 —7769.6  amp. ;  (b)  130.1  —747  volt.; 
(c)  2169.9  4-747  volt.  Use  the  inverse  operator  to  obtain  the 
current,  and  the  direct  operator  to  calculate  the  line  drop. 

PROB.  92.  A  voltage,  e  +  je' ,  is  impressed  across  the  impe- 
dances r^  -f  jxv  and  r2  +  jxz  in  parallel.  Find  the  total  current. 
Solution  :  The  total  conductance  is  g  —  rjz*  -f  rjz*,  and  the  total 
susceptance  is  b  =  xjz*  +  xjzf.  Hence,  the  current  i+ji'=. 
(e+je')  (g-jb)  =  (eg  4-  e'b}  +j(e'g  -  eb\ 

PROB.  93.  Extend  the  solution  of  problem  92  to  the  case 
where  more  than  two  impedances  are  in  parallel.  Ans.  i+ji'  — 
[e  1  rlz*  +  e'l  x\^\  +  j  \e'  2  r^  -el  x^  \ 

PROB.  94.  Two  impedances,  r^  -f  jxl  and  r2  4-  jxv  in  parallel, 
are  connected  in  series  with  a  third  impedance  r  -f  jx.  Show 
how  to  determine  the  total  voltage,  knowing  the  total  current 
i  +ji' ;  or,  how  to  find  the  expression  for  the  total  current  when 
the  total  voltage  e  +  je'  is  given. 

PROB.  95.  Show  how  to  solve  the  preceding  problem  when 
both  the  current  and  the  voltage  are  given,  but  either  the  impe- 
dance rx  +7X  or  the  impedance  r  +  jx  is  unknown. 

PROB.  96.  Four  given  resistances  and  reactances  are  connected 
as  shown  in  Fig.  9,  a  known  voltage  Ef  (generator  voltage)  being 

maintained   between  the  points  A  and  B.      Find  expressions, 
in  the  abbreviated  symbolic  notation,  for  the  current  through 


54  COMPLEX  QUANTITIES.  [CHAP.  5. 

the  impedance  r  +jx,  and  for  the  voltage  across  its  terminals.* 
Solution  :  Determine  an  impedance  Z3  as  the  sum  of  the  impe- 
dances r  +  jx  and  r2  +  jxv  Let  its  reciprocal,  or  the  equivalent 
admittance  be  K,,  and  the  admittance  of  the  circuit  rot  x0,  be  Y0. 
Then,  the  total  admittance  between  the  points  C  and  D  is  X3  -f 
Y0.  Find  the  corresponding  impedance  and  add  to  it  the  impe- 
dance of  the  part  rlt  xlt  of  the  circuit.  This  will  give  the  total 
impedance,  Z^t  between  A  and  B.  To  find  the  generator  current, 
multiply  the  given  voltage  by  the  inverse  operator,  Y^,  cor- 
responding to  this  impedance.  In  order  to  find  the  load  current 
and  voltage,  the  current  lost  in  the  circuit  r0,  x0  must  be  sub- 


?       ?                     a 

*y»    a 

X    o 

2 

Q 
f 

FIG.  9.     A  series-parallel  combination  of  impedances. 

tracted  from  the  generator  current,  and  the  voltage  drop  in  the 
part  AK  calculated.  This  is  done  exactly  as  in  problem  3, 
except  that  impedances  are  used  instead  of  resistances  in  calculat- 
ing the  drop,  and  the  admittance  Y0  is  used  instead  of  the  con- 
ductance g0  in  calculating  the  current  in  the  branch  r0,  x0.  In 
the  abbreviated  form  the  formulae  are  exactly  analogous  to  those 
in  problem  3,  except  for  the  notation.  However,  the  actual  sub- 
stitution of  complex  quantities  leads  to  long  and  complicated 
formulae,  and  is  of  no  advantage  for  general  purposes.  It  is 
preferable  to  perform  these  substitutions  on  a  numerical  example  ; 
see  next  problem. 


*This  diagram  of  connections  represents  what  is  known  as  the  equivalent 
transformer  diagram,  the  ratio  of  transformation  being  one  to  one.  The 
impedance  r  +jx  represents  the  load,  the  other  three  impedances  are 
intended  to  represent  the  various  losses  of  voltage  and  current  in  the  trans- 
former itself.  The  diagram  and  the  method  of  solution  are  analogous  to 
those  in  problem  3,  Chapter  I. 


CHAP.  5.]  COMPLEX  QUANTITIES.  55 

PROB.  97.  Apply  the  method  outlined  in  the  preceding  problem 
to  the  following  specific  case :  The  voltage  across  AB  is 
^=2740  +  7760  volt ;  r+jx  —  25  +710  ohm  ;  r^  +jxl=. 7  +ji. 2 

ohm  ;  r2  -\-  jx2  =  .8  +/ 1.5  ohm  ;  r0  =  760  ohm  ;  x0  =  540  ohm. 
Ans.  /=  94.86  —716. 94 amp.;  E=  2544  +/526 volt. 

PROB.  98.  Deduce  expression  (55 )  for  the  inverse  operator, 
without  the  use  of  the  direct  operator.  Solution  :  L,et  it  be  re- 
quired to  find  the  projections  of  the  current  vector,  knowing  the 
projections  of  the  voltage  vector  and  the  impedance  of  the  circuit. 
The  circuit  is  replaced  by  an  equivalent  parallel  combination  of 
a  conductance  £-  and  susceptance  b,  as  is  explained  in  Sec.  15. 
The  required  current  consists  of  two  components  :  A  working 
component,  in  phase  with  the  voltage,  equal  to  g  (e  +  je')\  and 
a  wattless  component,  lagging  behind  the  voltage  by  90  degrees, 
and  equal  to  b  (e  +  je'^jj ;  the  division  by/  representing  a  rota- 
tion in  the  negative  direction  by  90  degrees.  Adding  the  two 
component  vectors  of  current  and  factoring  out  e  +/<?'  we  get 
I  —  (<£"  +  %)  (e  4- /<?')>  or  multiplying  the  numerator  and  the 
denominator  of  the  term  b\j  by/, 

/=  Or -/*)O +/*'). 

Therefore,  g  —jb  is  the  operator  which  converts  the  vector  of 
the  voltage  into  that  of  the  current. 

19.  Power  and  Phase  Displacement  expressed  through 
Projections  of  Vectors.  I^et  an  alternator  supply  a  current 
/=  i  +ji'  at  a  voltage  E  =  e  +je',  and  let  it  be  required  to 

determine  the  power  output  of  the  generator.  The  expression 
for  the  average  power  is  P=  El cos  <£,  t where  <f>  is  the  phase  dis- 
placement between  /and  E  (see  Sec.  10).  The  angle  <£  is  the  dif- 
ference between  the  angles  Bl  and  02  which  the  vectors  E  and  / 
respectively  form  with  the  reference  axis.  Hence,  we  have 
P  =  EScos  <£  =  EScos  (O.  —  O^ 

=  Ecos  6l  -  /cos  e,  +  ^sin  Bl  -  /sin  02. ' 

Remembering  that  /scosfl^  /Tsin^,  etc.,  represent  the  projec- 
tions of  the  given  vectors  on  the  axes  of  co-ordinates  we  have 
simply 

P-ei+Sf (57) 


56  COMPLEX  QUANTITIES.  [CHAP.  5. 

Another  way  of  deducing  expression  (57)  is  to  resolve  the  given 
vectors  of  current  and  voltage  into  their  components  along  the 
axes  of  co-ordinates  and  to  consider  the  contribution  of  each  pro- 
jection to  the  total  power.  The  projections  e  and  i,  being  in 
phase,  give  the  power  ei.  Similarly,  the  vertical  projections,  e 
and  /',  give  the  power  e'i'  .  The  vertical  projection  i'  of  the 
current  gives  an  average  power  zero  with  the  horizontal  projec- 
tion e  of  the  voltage,  the  two  being  in  phase  quadrature.  For 
the  same  reason  the  average  power  resulting  from  e'  and  i  is 
equal  to  zero.  Thus,  ei  +  e'i'  represents  total  average  power. 

To  find  the  phase  displacement,  or  the  power  factor  of  the 
output,  we  write 

tan  *  =  tan  (^  -  *,)  =   tan  *'  ~  tan  *«.  , 
i  +  tan  0,  tan  02' 

or 


Knowing  tan  <£,  the  angle  itself  or  its  cosine  are  found  from  the 
trigonometric  tables  ;  or,  else,  the  power  factor  is  caculated  from 
the  formula  cos  <f>  =  (i  4-  tan2<£)~  l/2. 

Power  factor  can  be  also  determined  directly  from  the  ex- 
pression 

cos  <£  =  P\EI  =  (ei  +  e'i' 

but  the  calculation  is  more  involved  than  when  formula  (58)  is 
used. 

The  power  calculated  according  to  formula  (57)  sometimes 
comes  out  negative,  if  some  of  the  projections  of  E  and  /  are 

negative.  The  interpretation  is  that  in  this  case  the  phase  dis- 
placement between  the  current  and  the  voltage  is  over  90  degrees, 
so  that  power  is  being  supplied  to  the  machine,  instead  of  being 
delivered  by  it.  Tan  <f>  in  formula  (58)  can  also  be  negative, 
which  means  either  that  the  current  is  leading,  or  that  it  is  lag- 
ging by  an  angle  larger  than  90  degrees.  The  question  is  decided 
by  reference  to  the  sign  of  the  power. 

PROB.  99.  The  terminal  voltage  of  an  alternator  is  5370  -f/  735 
volts,  the  line  current  is  173  —  747  amps.     Calculate  the  output 


CHAP.  5.]  COMPLEX  QUANTITIES.  57 

of  the  machine  and  the  power  factor  of  the  load.     Ans.  894.5 
kilowatt.;  92  per  cent,  lagging. 

PROB.  100.  How  should  the  vertical  projection  of  the  current 
in  the  preceding  problem  be  modified  in  order  that  the  power 
become  zero?  Ans.  —  1264  amp. 

PROB.  101.  Let  the  line  current  in  problem  99  be  —  58  +/ 12 
amp.  Explain  the  negative  sign  of  power  and  the  plus  sign  of 
tan  <f>  ;  draw  the  vectors  of  the  current  and  of  the  voltage. 

PROB.  loia.  A  synchronous  machine  generates  a  voltage 
equal  to  2300 —750  volts,  and  supplies  a  current,  through  an 
impedance  5  -f  j  50  ohms,  to  another  synchronous  machine 
generating  a  counter-e.m.f.  of  2300  -f/5o  volts.  What  is  the 
power  output  of  the  first  machine  ?  Is  the  current  leading  or 
lagging?  Ans.  —  4.545  kw.  <£  =  173°  3'  lagging. 

PROB.  102.  A  current  of  350  —775  amp.  is  maintained  through 
an  impedance,  the  power  output  being  952  kw.  at  a  power  factor 
of  86  per  cent  lagging.  Find  the  voltage  across  the  impedance. 
Ans.  2930  +7987  volts.  Hint :  Solve  eqs.  (57)  and  (58)  together 
for  the  unknown  projections  e  and  e' . 

PROB.  103.  Solve  problem  102  by  calculating  the  value  of  the 
impedance,  and  multiplying  the  impedance  by  the  current. 
Hint :  power  =  Pr  ;  x  =  r  tan  <£. 

PROB.  104.  The  product  Pt  =  £7" sin  <£  is  called  the  reactive 
power  (wattless  power).  Find  its  expression  through  the  pro- 
jections of  the  vectors  of  the  voltage  and  the  current,  similar  to 
eq.  (57).  Ans.  Pt  =  e'i  —  ei' . 

PROB.  105.  Show  that  the  relation  P*  +  P*  =  P*  exists  be- 
tween the  true,  the  reactive  and  the  apparent  power ;  P&  =  EI 
is  called  the  apparent  power. 

PROB.  106.  Show  that  expression  (58)  can  be  also  written  in 
the  form  :  tan  <#>  =  PJP. 

PROB.  107.  Solve  problem  102  using  the  expressions  for  the 
reactive  power  derived  in  problems  104  and  106. 


CHAPTER  VI. 


THE  ELECTROSTATIC  CIRCUIT. 

20.  Physical  Concept  of  the  Electrostatic  Field.  The 
fundamental  phenomena  of  electrostatics  are  supposed  to  be 
known  from  physics.  The  purpose  of  this  chapter  is  to  give  a 
somewhat  different  interpretation  of  these  phenomena,  and  to 
deduce  numerical  relations  which  are  of  importance  in  electrical 
engineering.  The  reader  is  advised  to  consult  also  Chapter  XXI, 
on  Electrostatic  Capacity,  in  the  author's  Experimental  Electrical 
Engineering  (Chapter  VI  in  the  first  edition). 

Let  a  source  E  of  continuous  electromotive  force  (Fig.  10)  be 
connected  to  two  parallel  metallic  plates  A  and  B,  the  combina- 
tion of  which  is  commonly  known  as  a 
condenser.  The  plates  are  separated 
from  each  other  by  air,  or  by  some 
other  non-conducting  material.  When 
the  key  K  is  closed,  a  certain  quantity 
of  electricity,  Q,  flows  from  the  battery 
to  the  plate  A,  and  from  the  plate  B 
back  to  the  battery.  This  quantity 
can  be  measured  on  the  ballistic  gal- 
vanometer shown  in  the  circuit.  With- 
1 «.  vl' ''  /'  in  a  very  short  time  the  difference  of 

potential  between  the  plates   becomes 

FIG.  10.  A  plate  condenser    equal  to  that  of  the  battery,  so  that  the 
completing  a  direct-          flow  of  current  stops,      The  plates  are 

current  circuit.  ,,  .  ,    A      ,         .  ..        .  .  •'. 

then  said  to  be  charged  with  certain 

quantities  of  electricity,  +  Q  and  —  Q, 

From  a  modern  point  of  view,  it  is  preferable  to  consider 
electricity  as  an  incompressible  fluid.  Therefore,  no  charges  are 
accumulated  on  the  plates,  but  a  quantity  of  electricity  Q  is  dis- 
placed in  the  circuit,  including  the  layer  of  the  insulation  or 
dielectric  between  the  plates.  This  displacement  is  accompanied 
in  the  dielectric  by  a  stress,  similar  in  some  respects  to  a 
mechanical  stress  in  an  elastic  body.  The  direction  of  the 
electric  stress  and  of  the  lines  of  displacement  of  electricity  are 


jBall               £ 

— 

Cffi*-  ,"•*—>, 

1          \'^ 

r       > 
£J 

N        1           ' 
// 

K,j 

/f 

^~§i 

# 

+£ 

'-!-'- 

-a 

Condenser 

•'x; 

:^<s 

/  //-. 

•^x 

ft; 

~- 

f\x 

CHAP.  6.] 


THE  ELECTROSTATIC  CIRCUIT, 


59 


shown  in  the  figure  by  dotted  lines.  When  the  key  is  opened, 
the  condenser  remains  charged,  since  the  stress  and  the  displace- 
ment can  be  equalized  only  in  a  closed  circuit.  To  discharge 
the  condenser  its  plates  must  be  connected  by  a  conductor. 
Then  the  deflection  of  the  ballistic  galvanometer  is  equal  and 
opposite  to  that  during  the  charge,  and  the  electric  energy  stored 
in  the  condenser  is  dissipated  in  the  form  of  heat  by  the  current. 
The  difference  between  a  dielectric  and  a  conductor  is  that  the 
resistance  of  the  former  to  the  passage  of  electricity  is  of  an 
elastic  nature  ;  that  is  to  say,  the  stress  can  be  relieved  and  the 
stored  energy  can  be  returned  to  the  circuit.  On  the  contrary, 
the  resistance  to  the  flow  of  electricity  in  a  conductor  has  the 
nature  of  friction.  The  energy  spent  is  converted  into  Joulean 
heat  and  cannot  be  restored. 


FIG.  ii.  Electric  stress  and  displacement  in  a  dielectric, 

represented  by  filaments  of  positive  and  negative 

electricity,  shearing  past  one  another. 

When  the  voltage  E  is  sufficiently  high,  the  displacement  of 
electricity  in  the  dielectric  reaches  a  limit  at  which  the  material 
is  "broken  down",  and  a  disruptive  discharge,  or  a  spark, 
occurs  between  the  edges  of  the  plates.  With  solid  and  liquid 
insulating  materials,  such  as  glass,  oil,  mica,  etc.,  larger  charges 
or  displacements  Q  are  obtained  in  the  same  condenser,  with  the 
same  voltage  E. 

These  and  other  phenomena  of  electrostatics  are  accounted  for 
by  assuming  a  certain  hypothetical  structure  of  dielectrics. 
Namely,  a  dielectric  is  supposed  to  contain  in  its  natural,  un- 
electrified  state  both  positive  and  negative  electricity,  combined 
so  as  neutralize  each  other.  The  process  of  electrification  (Fig. 
n)  consists  in  displacing  positive  electricity  towards  one 


60  THE  ELECTROSTATIC  CIRCUIT.  [CHAP.  6. 

electrode  or  plate,  and  negative  electricity  towards  the  other 
plate.  This  separation,  caused  by  the  applied  electromotive 
force,  produces  a  stress  in  the  dielectric.  While  the  actual 
mechanism  of  the  phenomenon  is  unknown,  Fig.  n  give  a  simple 
and  possible  picture  of  it. 

The  positive  and  the  negative  electricity  are  represented  by 
alternate  threads  or  filaments.  The  electromotive  force,  accord- 
ing to  this  analogy,  pulls  positive  threads  in  one  direction,  and 
negative  ones  in  the  opposite  direction.  The  connection  between 
adjacent  threads  must  be  assumed  to  be  of  an  elastic  nature,  so 
that  this  pull  produces  a  shearing  stress  along  the  periphery  of 
the  threads  ;  this  stress  opposes  their  displacement.  The  dis- 
placement reaches  its  limit  when  the  shearing  stress  just  balances 
the  applied  electromotive  force.  The  same  filaments,  or  some- 
thing corresponding  to  them  (for  instance  paths  of  electrons), 
must  be  assumed  to  continue  in  the  conductors,  but  there  the 
resistance  to  the  motion  is  of  the  nature  of  friction.  This  friction, 
as  in  ordinary  fluids,  must  be  supposed  to  be  proportional  to  the 
velocity  of  motion.  Therefore,  it  is  manifested  only  when 
electricity  is  in  motion,  in  other  words,  when  a  current  is  flowing. 
This  assumption  is  consistent  with  Ohm's  law  for  conductors, 
and  hence  may  be  adopted. 

A  stress,  and  consequently  a  difference  of  potential,  exists  in 
a  dielectric  when  the  threads  have  been  displaced  from  their 
natural  relative  position,  though  the  threads  themselves  may  be 
stationary  in  the  new  position.  The  stress  is  relieved  by  remov- 
ing the  electromotive  force  and  by  providing  a  closed  path  of  low 
resistance,  through  which  the  threads  can  again  be  relatively 
displaced  into  their  natural  position.  What  according  to  the  old 
theory  is  considered  as  charges  on  condenser  plates,  is  but  a  dis- 
placement in  the  intervening  dielectric.  The  plates  are  only  the 
boundaries  of  this  dielectric,  on  which  boundaries  the  phenome- 
non apparently  takes  place.  When  a  current  is  flowing  through 
the  metallic  part  of  the  circuit  an  equal  displacement  current  is 
established  in  the  dielectric,  which  thus  completes  the  circuit. 

When  an  alternating  voltage  is  applied  at  the  terminals  of  a 
condenser,  the  displacement  in  the  dielectric  changes  continually 
in  its  magnitude  and  direction  ;  consequently,  it  gives  rise  to  an 
alternating  current  in  the  metallic  part  of  the  circuit.  This  is 


CHAP.  6.]  THE  ELECTROSTATIC  CIRCUIT.  6 1 

called  the  charging  or  the  capacity  current.  This  current  leads 
the  alternating  voltage  in  phase  by  90  degrees,  as  may  be  seen 
from  the  following  considerations  :  When  the  voltage  has  reached 
its  maximum,  the  charging  current  is  zero,  because  at  the  crest 
of  the  wave  the  voltage  and  the  displacement  remain  practically 
constant  for  a  short  period  of  time.  As  soon  as  the  voltage 
begins  to  decrease,  the  current  begins  to  flow  in  the  direction 
opposite  to  that  of  the  applied  voltage,  because  the  elastic  reac- 
tion of  the  dielectric  is  now  larger  than  the  applied  electromotive 
force.  At  any  instant  the  current,  or  the  rate  of  flow  of  elec- 
tricity, is  proportional  to  the  rate  of  change  of  the  applied 
voltage.  But  if  the  applied  voltage  varies  according  to  the  sine 
law,  the  rate  of  variation  is  also  represented  by  a  sine  function, 
differing  in  phase  by  90  degrees  from  the  original  function 
[d  (sin  x)jdx  =  cos  x  —  sin  (90  -f  x)~\ .  It  is  well  worth  the 
student's  while  to  think  over  this  phase  relation  between  the 
charging  current  and  the  voltage  more  in  detail.  That  there 
must  be  a  displacement  of  90  degrees  between  the  voltage  and 
the  current  follows  also  directly  from  the  assumed  elastic  structure 
of  the  dielectric.  Namely,  the  energy  is  supposed  to  be  periodi- 
cally stored  in  the  dielectric  and  given  up  again  without  any  loss. 
Hence,  the  average  power  must  be  zero,  and  the  current  must 
be  wattless.  See  also  Art.  24  below. 

A  hydraulic  analogy  shown  in  Fig.  12  may  assist  the  student 
in   the   understanding   of   the   mechanism   of   the   electrostatic 

circuit.  A  is  a  pump  which  cor- 
responds to  the  source  of  electromotive 
force  in  Fig.  10.  The  pipes  B  and  C 
represent  the  leads  to  the  condenser, 
or  the  metallic  parts  of  the  circuit. 
The  cylinder  D  corresponds  to  the 
condenser,  and  the  elastic  partition  K 
is  analogous  to  the  dielectric.  Let  the 

pipes  and  the  cylinders  be  filled  with 
FlG.  12.  A  hydraulic  analogue  ,  ,         .         .  ..... 

of  an  electrostatic  circuit.       Water'  and  let  the  P1StOn  ln  A  be  m  lts 

middle  position,  the  partition  K  being 

not  stressed.  Let  the  stop-cock  Mbe  open,  and  the  stop-cock  N 
be  closed.  When  a  pull  to  the  left  is  exerted  upon  the  piston  rod 
and  it  is  made  to  move,  the  water  in  the  system  is  displaced,  and 


62  THE  ELECTROSTATIC  CIRCUIT.  [CHAP.  6. 

the  elastic  partition  K  is  strained,  as  is  shown  in  the  figure.* 
With  a  given  pull,  or  a  given  electromotive  force,  the  movement 
stops  when  the  pull  is  balanced  by  the  elastic  reaction  of  the 
partition.  The  charge,  or  the  total  displacement,  is  represented 
by  the  amount  of  the  water  shifted  ;  it  can  be  measured  on  the 
water  meter  W,  which  thus  takes  the  place  of  the  ballistic 
galvanometer. 

If  the  pipes  are  frictionless,  the  analogy  can  be  followed  still 
further  ;  namely,  the  phase  difference  in  time  between  the  pull 
and  the  velocity  of  the  water  is  90  degrees,  the  latter  leading  the 
pull.  Assuming  the  motion  of  the  piston  to  be  harmonic,  the 
velocity  of  the  flow  of  water  is  at  its  maximum,  when  the  piston 
is  at  the  center  of  its  stroke.  The  required  pull  is  equal  to  zero 
at  this  moment,  because  the  elastic  partition  is  in  its  middle, 
unstrained  position.  At  the  end  of  the  stroke,  the  velocity  is 
zero,  but  the  pull  is  at  its  maximum,  because  the  partition  is 
strained  to  its  extreme  position,  and  exerts  its  maximum  elastic 
reaction. 

Substituting  another  partition,  made  of  a  more  yielding  ma- 
terial (material  possessing  higher  permittivity)  a  larger  dis- 
placement is  produced  with  the  same  pull ;  this  corresponds  to 
the  case  when  some  solid  or  liquid  dielectric  is  substituted  for 
the  air.  The  quantit}'  of  electricity  displaced  per  unit  pull  (that 
is  per  unit  voltage)  is  called  the  capacity  of  the  condenser. 
Thus  we  see  that  the  capacity  of  a  condenser  is  increased  by  the 
substitution  of  a  material  possessing  higher  permittivity. 

Closing  the  stop-cock  M  corresponds  to  breaking  the  electrical 
circuit  of  the  condenser.  It  will  be  seen  that  the  condenser 
remains  charged.  To  discharge  the  condenser,  the  stop-cock  N 
must  be  opened  ;  this  equalizes  ,the  pressure  on  both  sides  of  the 
elastic  partition.  Since  water  possesses  some  inertia  the  parti- 
tion does  not  stop  in  its  middle  position  during  the  discharge, 
but  the  momentum  of  the  water  carries  it  beyond  the  center.  The 
electromagnetic  inertia  of  the  electric  current  produces  a  similar 
effect,  and  we  thus  have  an  explanation  of  the  oscilatory  char- 
acter of  the  electric  discharge.  During  this  discharge  the  energy 


*  The  diameter  of  the  cylinder  K  must  be  conceived  to  be  many  times 
larger  than  that  of  the  pump  A. 


CHAP.  6.]  THE  ELECTROSTATIC  CIRCUIT.  63 

is  alternately  transformed  into  the  potential  energy  of  the  die- 
lectric stress,  and  into  the  kinetic  energy  of  the  magnetic  field. 
The  oscillations  of  the  partition  are  gradually  damped  out  by  the 
frictional  resistance  of  the  pipes.  In  the  electric  circuit  oscilla- 
tions are  damped  by  the  ohmic  resistance  in  the  conducting  parts 
of  the  circuit. 

The  student  can  follow  this  analogy  still  further  and  explain 
the  natural  period  of  vibration,  current  and  voltage  resonance, 
and  the  effect  of  a  resistance  in  series  and  in  parallel  with  a 
condenser,  etc. 

21.  Dielectric  Flux  Density  and  Electrostatic  Stress 
(Voltage  Gradient).  Referring  again  to  the  uniform  electro- 
static field  in  Fig.  10,  consider  a  cube  in  the  dielectric,  one  square 
centimeter  in  cross-section,  and  one  centimeter  long  in  the  di- 
rection of  the  lines  of  force.  L,et  a  quantity  of  electricity  Q  be 
supplied  by  the  battery,  as  shown  by  the  ballistic  galvanometer ; 
then  the  same  quantity  of  electricity  must  be  displaced  in  the 
dielectric.  Neglecting  a  small  displacement  at  the  edges  and  at 
the  outside  surfaces  of  the  plates,  it  may  be  said  that  the  whole 
quantity  Q  is  uniformly  displaced  between  the  plates.  There- 
fore, if  the  area  of  each  plate  is  equal  to  A  square  centimeters, 
the  displacement  through  the  cube  under  consideration  is  equal  to 

D=QIA.     .......     (59) 

Since  Q  is  the  total  electrostatic  flux,  D  is  naturally  called  the 
dielectric  flux  density.  If  Q  is  measured  in  coulombs,  D  is  ex- 
pressed in  coulombs  per  square  centimeter.  In  practice  Q  is 
measured  in  microcoulombs,  and  D  is  expressed  in  microcoulombs 
per  square  centimeter,  or  per  square  inch.  It  will  be  remem- 
bered that  a  coulomb  is  the  quantity  of  electricity  which  a  cur- 
rent of  one  ampere  supplies  during  one  second.  One  micro- 
coulomb  is  one  millionth  part  of  a  coulomb. 

By  analogy  with  the  shearing  stress  in  the  theory  of  elasticity, 
the  stress  in  the  dielectric  (Fig.  n)  can  be  defined  as  the  sum  of 
the  shearing  forces  between  the  positive  and  the  negative  fila- 
ments within  a  unit  cube.  When  the  condenser  is  fully  charged, 
the  sum  of  these  shearing  forces  is  balanced  by  the  electromotive 
force  between  the  plates.  Thus,  if  the  distance  between  the 
plates  is  /  centimeters  and  the  voltage  is  E,  we  have  E  —  I  •  F, 


64 


THE  ELECTROSTATIC  CIRCUIT. 


[CHAP.  6. 


where  F  is  the  electrostatic  stress,  or  the  force  per  unit  length 
of  the  unit  filament  (the  force  per  unit  cube).  In  other  words, 

F=E\l (60) 

or,  the  electrostatic  stress  in  a  uniform  field  is  equal  to  the  voltage 
gradient  per  unit  length  of  the  field.  F  is  expressed  in  volts  (or 
in  kilovolts)  per  centimeter,  or  per  inch.* 


FIG.  13.  A  non-uniform  electrostatic  field,  represented  by 
lines  of  displacement  and  by  equipotential  surfaces. 

Equations  (59)  and  (60)  connect  two  quantities,  the  voltage 
E  and  the  charge  Q,  with  the  flux  density  D  and  the  stress  F. 
The  first  two  quantities  characterize  the  circuit  as  a  whole,  the 
two  latter  characterize  the  dielectric  proper.  It  will  be  seen 
that  these  equations  are  analogous  to  the  eqs.  (n)  and  (12)  in 
Chapter  I.  The  expression  ''electric  intensity"  used  there  is 
the  same  as  the  expression  "stress  "  used  here.  It  is  also  proper 
to  call  .Fthe  voltage  gradient. 


*The  voltage  E  is  the  cause  of  a  state  of  stress  in  a  dielectric  of  thickness 
/.  Since,  by  supposition  this  state  is  uniform  throughout,  it  takes  a  voltage 
E\l  to  produce  this  stress  per  unit  thickness  of  dielectric.  This  quantity  is 
denoted  as  the  stress  F.  With  this  interpretation  it  is  not  necessary  to 
define  or  describe  by  analogies  the  nature  of  the  electric  stress. 


CHAP.  6.]  THE  ELECTROSTATIC  CIRCUIT.  65 

When  the  field  is  non-uniform,  as  in  Fig.  13,  an  infinitesimal 
parallelepiped  mnpq  must  be  considered,  instead  of  a  unit  cube. 
Let  the  cross-section  mp  of  this  parallelepiped,  perpendicular  to 
the  direction  of  the  field,  be  dS,  and  the  length  mn  in  the  direcr 
tion  of  the  field  equal  dl.  Let  the  displacement  through  the 
parallelepiped  be  dQ.  Then  the  dielectric  flux  density,  or  the 
displacement  per  unit  area,  is 

D=dQldS.     ......      (61) 

Again,  let  the  voltage  between  the  opposite  sides  mp  and  nq  of 
the  parallelopiped,  in  the  direction  of  the  field,  be  dE.  Then, 
the  stress,  or  the  voltage  gradient,  at  the  center  of  the  parallelo- 
piped is  equal  to 

F=dE\dl.     .     .     .      .      .      .      (62) 

Equations  (61)  and  (62)  must  be  used  in  place  of  eqs.  (59)  and 
(60)  when  the  field  is  non-uniform.  They  give  D  and  F  as 
functions  of  Q  and  E.  If  it  is  desired  to  express  Q  and  E  in 
terms  of  D  and  Ft  the  foregoing  equations  must  be  integrated. 
Integrating  eq.  (61)  we  get 


Q  =  C 

V    o 


•  <tS,     .     .     .     .     .     .      (63) 


where  the  integration  has  to  be  extended  over  a  complete  closed 
equipotential  surface,  such  as  XY,  surrounding  one  of  the  con- 
denser plates.  This  follows  from  the  fact  that  dS  is  selected  in 
each  elementary  parallelopiped  in  a  plane  perpendicular  to  the 
displacement,  and  hence  lies  in  an  equipotential  surface.  Inte- 
grating eq.  (62)  we  obtain 


=/ 

*/    n 


F-  dl,      .     .     .     .     .     .      (64) 


where  the  integration  is  to  be  performed  along  a  line  of  force, 
such  as  HH1 ',  between  the  positive  and  negative  plates  of  the 
condenser. 

The  last  two  relations  are  expressed  in  words  by  saying  that 
the  total  displacement  Q  is  a  surface  integral  of  the  dielectric 
flux  density,  while  the  voltage  E  is  a  line  integral  of  the  die- 
lectric stress.  A  little  consideration  will  show  that  these  relations 
are  almost  self  evident  ;  at  least  that  they  follow  directly  from 


66  THE  ELECTROSTATIC  CIRCUIT.  [CHAP.  6. 

the  definitions  of  the  quantities  concerned,  and  from  the  assumed 
structure  and  properties  of  dielectrics.  Compare  also  with  Article 
5  in  Chapter  I. 

PROB.  108.  A  condenser  (Fig.  10)  consists  of  two  metal  plates, 
50  X  70  cm.  each,  in  contact  with  a  glass  plate  3  mm.  thick 
between  them.  When  a  continuous  voltage  of  2400  v.  is  applied 
to  the  condenser,  the  ballistic  galvanometer  shows  a  charge  of 
17.1  microcoulombs.  What  is  the  dielectric  flux  density  and  the 
stress  in  the  glass?  Ans.  D  =  17.1  /(5o  x  70)  =  4.885  X  io~s 
me.  /cm.2;  F=  2.4.1.3  =  8kv./cm. 

PROB.  109.  What  is  the  flux  density  and  the  voltage  gradient 
in  the  preceding  problem,  expressed  in  the  English  system? 
Ans.  Z)=3i.52  x  i  o-3  me.  /inch2;  F—  20.32  kv./inch. 

PROB.  no.  A  certain  kind  of  oil-cloth  insulation  is  broken 
down  between  two  plates  at  a  voltage  gradient  of  about  20  kv. 
per  mm.  (maximum  instantaneous  value).  What  should  be  the 
thickness  of  this  insulation  between  two  plates  to  stand  45 
effective  kilovolts,  at  a  factor  of  safety  of  5  ?  Ans.  About  5/8 
inch. 

PROB.  in.  A  single-core  cable  (Fig.  14)  receives  a  charge  of 
.0019  coulombs  per  mile  when  a  continuous  voltage  of  12  kv.  is 
applied  between  the  core  and  the  sheath- 
ing.    The  core  consists  of  No.  4  solid  con- 
ductor B.  &  S.,  the  insulation  is  3/8  inch 
thick.      Determine  the  law  according  to 
which    the  dielectric  flux  density    varies 
within   the   insulation,    and    the  extreme 
values  of  the  density.     Ans.   Equilateral 
hyperbola,     D  =  4.77  X   icr*lx    me.    per 
FIG.  14.  Cross-section  of  square  inch,  where  x  is  the  radius  of  the 
a  single-core,  lead-       layer  under  consideration,  in  inches  ;  DmM 
covered  cable.  .   D      =    QI 


PROB.  112.  The  cable  insulation  in  the  preceding  problem 
breaks  down  at  a  maximum  instantaneous  flux  density  of  .4 
mc./inch2.  At  what  alternating  voltage  between  the  core  and 
the  sheathing  does  this  occur,  and  which  layer  of  insulation 
breaks  down  first  ?  Ans.  At  about  73  effective  kilovolts,  near 
the  core.  Hint  :  Flux  density  is  proportional  to  the  voltage. 


CHAP.  6.]  THE  ELECTROSTATIC  CIRCUIT.  67 

22.  Permittivity  and  Dielectric  Strength.  As  long  as  an 
insulating  material  is  stressed  well  below  the  point  at  which  a 
disruptive  discharge  occurs,  the  dielectric  flux  density  is  propor- 
tional to  the  stress,  that  is  to  say 

D  =  KF. (65) 

Here  K  is  a  coefficient  of  proportionality  called  the  permittivity 
of  the  dielectric.  This  law  expresses  the  fact  that  the  effect  is 
proportional  to  the  cause.  It  is  analogous  to  Hooke's  law  for 
elastic  materials  :  displacement  corresponds  to  strain,  and  I/K 
corresponds  to  the  modulus  of  elasticity.  Eq.  (65)  is  also 
analogous  to  Ohm's  law,  in  the  form  given  by  eq.  (10).  To 
conductivity  y  corresponds  here  the  permittivity  K.  Conductivity 
indicates  the  ease  with  which  electricity  can  flow  through  a  metal 
or  imperfect  insulator  ;  permittivity  shows  the  ease  with  which 
electricity  can  be  displaced  through  a  dielectric.  One  indicates 
frictional  resistance  to  motion,  the  other  an  elastic  resilience. 
For  air  the  permittivity  in  the  metric  system  is  equal  to 

Ka  =  .08842  X  lo"6     .     .*    .     ,     .,    (66) 

microcoulombs  per  volt,  per  square  centimeter  cross- section,  per 
centimeter  length.  In  other  words,  in  an  air  condenser,  with 
the  distance  between  the  plates  equal  to  one  centimeter,  the  dis- 
placement per  one  square  centimeter  of  cross-section  of  the 
dielectric  is  equal  to  .08842  x  icr"6  microcoulombs,  when  the  dif- 
ference of  potential  between  the  plates  is  one  volt.  The  dimen- 
sions of  the  plates  are  assumed  to  be  very  large  as  compared  to 
the  distance  between  them,  so  that  the  influence  of  the  edges 
can  be  neglected.  See  Note  at  the  end  of  the  chapter. 
The  value  of  permittivity  of  air  in  the  English  system  is 

Ka=  .2244  x  io~6  .     .     .    .    .     .     (67) 

in  microcoulombs  per  one  square  inch  of  the  area  of  the  dielec- 
tric, per  volt,  per  inch  of  the  length  of  the  lines  of  force. 

For  other  substances,  it  is  convenient  to  use  relative  values  of 
permittivities,  referred  to  that  of  air  as  unity.*  In  this  way, 
the  relative  properties  of  various  dielectrics  are  made  more  evi- 
dent, and  the  necessity  is  obviated  for  tabulating  very  small 

*The  older  name  for  relative  permittivity  is  specific  inductive  capacity. 


68 


THE  ELECTROSTATIC  CIRCUIT. 


[CHAP.  6. 


quantities,  like  those  in  eqs.  (66)  and  (67).  The  values  of 
relative  permittivities  for  some  materials  are  given  in  the  table 
below,  in  the  second  column. 


Substance. 

Relative  permit 
tivity,  or  specific 
inductive  capac- 
ity K. 

Rupturing  volt- 
age gradient 
Fmia,  in  eff.  kv. 
per  mm. 

Rupturing  values  of 
dielectric  flux  densi- 
ties, Anax,  in  eff  me. 
per  sq  cm. 

Air 

j 

2.7 

.0024 

Glass,  different  kinds  _ 
Mica,  natural  and  built 
up      

3-8 
5—8 

6-18 
17  —  28 

.016  —  .056 
.075  —  .2OO 

Porcelain,  B.C.     

5.3 

Porcelain,  A.C.__ 

4.4 

9  —  16 

.035  —  .062 

Rubber,  pure 

2.2 

20  —  25 

.039  —  .049 

Rubber,  vulcanized  
Transformer  oil 

2.7 
2  —  2.2 

15—20 
10 

.036  —  .048 
.Ol8  —  .019 

Vacuum  

•99 

NOTE  :  The  data  given  in  the  table  indicate  the  order  of  magnitude  rather 
than  accurate  values.  The  last  two  columns  contain  the  effective  values  of 
voltage  gradient  and  of  displacement,  under  the  supposition  of  a  sine-wave 
law  of  variation  of  the  applied  voltage. 

It  will  be  seen  from  the  table,  that  the  permittivities  of  solid 
and  liquid  dielectrics  are  larger  than  those  of  air  ;  in  other  words, 
they  are  more  yielding  to  the  electric  stress  than  the  air.  This 
does  not  mean,  however,  that  they  break  down  at  a  lower  voltage 
gradient  than  the  air.  On  the  contrary,  the  third  and  the  fourth 
columns  show  that  the  dielectrics  commonly  used  in  electrical 
engineering  are  considerably  stronger  electrically  than  the  air, 
in  that  they  can  stand  several  times  the  displacement  at  which 
the  air  breaks  down. 

The  above-mentioned  fact  indicates  that  the  permittivity  alone 
is  not  sufficient  to  characterize  an  insulating  material  for  practical 
uses,  but  that  its  "dielectric  strength"  must  also  be  known. 
There  does  not  seem  to  be  any  relation  between  these  two  co- 
efficients :  One  indicates  the  elasticity  of  the  material,  the  other 
its  ultimate  strength.  In  this  respect,  they  are  analogous  to  the 
modulus  of  elasticity  and  to  the  rupturing  load  in  the  mechanics 
of  materials.  Air,  from  an  electrical  point  of  view,  may  be  said 
to  be  a  material  of  great  stiffness,  but  one  which  breaks  at  a 
comparatively  small  elongation.  On  the  contrary,  mica  is  com- 
paratively yielding,  but  can  stand  a  very  large  elongation  before 
it  is  ruptured  ;  so  that,  in  spite  of  larger  permittivity,  a  much 


CHAP.  6.]  THE  ELECTROSTATIC  CIRCUIT.  69 

higher  stress  is  required  to  rupture  mica  than  air.  The  student 
is  advised  to  make  clear  to  himself  these  two  separate  properties 
of  dielectrics :  A  rational  design  of  high-tension  insulation 
'depends  essentially  upon  a  distinct  understanding  of  these 
properties. 

Dielectric  strength  is  properly  given  as  the  critical  flux  density, 
Auax>  but  for  practical  purposes  it  is  more  convenient  to  express 
it  as  the  voltage  gradient,  Fmwi,  at  which  the  dielectric  is  broken 
down.  When  a  dielectric  is  used  in  the  form  of  thin  sheets 
having  a  comparatively  large  radius  of  curvature,  the  flux 
density,  and,  consequently,  the  voltage  gradient  are  uniform 
throughout,  so  that  Fm&x  =  F^.  When,  however,  the  dielectric  is 
thick  as  compared  to  its  radius  of  curvature,  as  for  instance  in 
high-tension  machines,  or  when  air  or  oil  are  tested  between  two 
spherical  terminals,  the  use  of  the  average  voltage  gradient 
Fwe  =  E\l  leads  to  wrong  results.  The  only  proper  way  in  this 
case  is  to  calculate  the  dielectric  flux  density  where  it  is  at  a 
maximum,  and  to  see  that  this  density,  /?max,  and  the  correspond- 
ing value  of  the  voltage  gradient,  /rmax  =  DmJ*,  do  not  exceed 
the  critical  values,  determined  from  previous  tests. 

PROB.  113.  Show  how  the  value  of  /ca  in  eq.  (67)  is  derived 
from  that  in  eq.  (66).  Ans.  .08842  X  lo"6  x  (2.54)8/2.54  = 
.2244  X  io~6. 

PROB.  114.  Show  how  the  values  in  the  last  column  of  the 
table  are  derived  from  those  in  the  two  preceding  columns. 
Ans.  Z>max  =  .08842  KFmn  x  io-2. 

PROB.  115.  What  is  the  relative  permittivity  (specific  inductive 
capacity)  of  the  glass  in  problem  108?  Ans.  6.9. 

PROB.  116.  A  certain  material  stood  about  82  kv.  in  a  layer 
3.7  mm.  thick.  What  voltage  gradient  can  be  allowed  in  this 
material  at  a  factor  of  safety  equal  to  2  ?  Ans.  n  kv.  per  mm. 
thickness. 

23.  Permittance,  or  Electrostatic  Capacity.  The  total 
displacement  produced  in  the  dielectric  of  a  given  condenser  is 
proportional  to  the  voltage  at  the  terminals  of  the  condenser,  be- 
cause voltage  and  displacement  appear  as  cause  and  effect. 
Hence,  we  have  the  relation 

Q=CE, (68) 


70  THE  ELECTROSTATIC  CIRCUIT.  [CHAP.  6. 

where  the  coefficient  of  proportionality  C  is  properly  called  the 
permittance  of  the  condenser.  The  larger  C  the  larger  the  total 
displacement  Q  for  a  given  voltage,  or  the  more  yielding  is  the 
condenser.  The  older  name  for  permittance  is  electrostatic 
capacity.  The  name  permittance  is  much  more  preferable,  for 
the  following  reasons  :  ( i )  It  suggests  the  true  action,  of  the 
dielectric,  while  the  word  capacity  was  intended  to  characterize 
the  metallic  plates  of  the  condenser  according  to  the  old  view, 
that  is  to  say,  with  regard  to  their  ability  to  store  or  to 
"condense"  electricity.*  (2)  Permittivity  and  permittance 
stand  in  the  same  relation  to  each  other  as  resistivity  and  re- 
sistance, or  as  permeability  and  permeance,  conductivity  and 
conductance,  etc.,  so  that  a  uniformity  of  notation  is  preserved. 

The  unit  of  permittance  or  capacity  in  the  volt-ampere-ohm 
system  is  the  farad.  A  condenser  is  said  to  have  its  permittance 
equal  to  one  farad  when  one  coulomb  of  electricity  is  displaced 
in  its  dielectric  per  volt  of  pressure  between  its  terminals.  The 
farad  is  by  far  too  large  a  unit  for  practical  purposes  ;  therefore, 
permittance  is  usually  measured  in  microfarads.  The  permit- 
tivity K  which  has  been  defined  above  in  terms  of  microcoulombs 
per  volt,  per  cubic  unit  of  dielectric,  now  can  be  more  simply 
defined  as  microfarads  per  cubic  centimeter,  or  per  cubic  inch. 

When  the  lines  of  force  in  a  prismatic  slab  of  dielectric  are 
parallel  to  each  other  and  parallel  to  one  of  the  principal  dimen- 
sions of  the  slab,  its  permittance  is  expressed  by  a  formula  similar 
to  that  for  the  conductance  of  a  prismatic  conductor.  Namely, 

C=KAIl.     ..;..._..    .     .     ,     .    (69) 

This  follows  directly  from  the  assumed  structure  of  the  dielec- 
tric. The  reasoning  is  the  same  as  in  the  deduction  of  the 
corresponding  expression  for  the  conductance:  When  A  =  i, 
and  /  =  i ,  the  permittance  is  by  definition  equal  to  permittivity 
K.  If  the  cross-section  is  increased  A  times,  as  many  times  more 
electricity  is  displaced  with  the  same  voltage  between  the  plates. 
If  the  length  is  increased  /  times,  the  voltage  gradient  E\l  be- 


*An  incompressible  substance  like  electricity  evidently  cannot  be  con- 
densed. To  complete  the  rational  nomenclature,  the  word  "condenser" 
should  be  replaced  by  "permittor"  or  "elastor",  to  correspond  with 
' '  conductor ' '  and  ' '  resistor. ' ' 


CHAP.  6.]  THE  ELECTROSTATIC  CIRCUIT,  71 

comes  /  times  smaller,  and  consequently  /  times  less  electricity  is 
displaced  per  unit  area. 

Applying  the  foregoing  formula  to  the  condenser  in  problem 
108,  its  permittance  is  found  to  be  equal  to 

C—  .08842  X  io~6  x  6.9  X  50  X  70/.3  =  .00712  mf., 
where  the  relative  permittivity  of  the  glass,  K '—  6.9,  is  supposed 
to  be  given. 

When  the  dielectric  in  a  condenser  is  not  prismatic  in  form, 
which  is  mostly  the  case,  its  permittance  is  determined  by  divid- 
ing the  dielectric  into  infinitesimal  layers,  or  into  a  large  number 
of  very  thin  layers  and  threads  (Fig.  13).  The  permittance  of 
'each  element  can  be  expressed  by  the  formula  (69).  The  per- 
mittance of  the  whole  dielectric  is  obtained  by  combining  the 
layers  and  the  threads  in  series  and  in  parallel,  as  the  case  may 
be.  The  procedure  is  analogous  to  that  of  finding  the  resis- 
tance of  a  conductor  of  variable  cross-section  (see  Sec.  5). 

This  leads  to  the  question  of  combining  permittances,  or 
capacities,  in  series  and  in  parallel.  When  two -or  more  permit- 
tances are  connected  in  parallel  between  the  same  electrodes,  the 
combined,  or  the  equivalent  permittance  is  equal  to  the  sum  of 
the  component  permittances.  The  proof  is  exactly  the  same  as 
in  the  case  of  conductances  in  parallel,  Sec.  2  ;  thus,  we  have 

C.,=  SC. (70) 

When  permittances  are  connected  in  series,  the  equivalent  per- 
mittance is  smaller  than  any  one  of  the  component  permittances, 
because  the  voltage  drop  is  distributed  over  a  greater  length, 
and  the  stresses  become  smaller.  AS  in  the  case  of  conductances, 
we  have 

C-'  =  2C-.     .    .    .    .     .    .   (71) 

For  a  formal  proof  of  eqs.  (70)  and  (71)  see  Art.  430  (Art.  121 
in  the  first  edition)  of  the  author's  Experimental  Electrical 
Engineering.* 


*When  permittances  are  connected  in  series,  it  is  more  convenient  to 
treat  the  reciprocal  C~ l  of  each  permittance  as  a  new  physical  quantity, 
which  can  be  properly  called  the  elastance  of  the  dielectric.  Then  eq.  (71) 
simply  expresses  that  the  equivalent  elastance  is  equal  to  the  sum  of  the 
elastances  in  series  ;  this  is  analogous  to  the  addition  of  resistances  in  series. 
Conductance  and  permittance  show  the  ease  with  which  electricity  can  flow 


72  THE  ELECTROSTATIC  CIRCUIT.  [CHAP.  6. 

PROB.  117.  A  condenser  which  has  a  permittance  of  10  mf.  is 
connected  to  a  direct-current  magneto,  the  speed  of  which  is 
increased  at  a  uniform  rate,  so  that  the  voltage  rises  at  a  rate  of 
1.7  volts  per  sec.  Calculate  the  charging  current.  Ans.  17 
microamperes.  NOTE.  This  is  the  principle  of  an  apparatus 
used  for  measuring  the  acceleration  of  railway  trains. 

PROB.  118.  A  .5  mf.  mica  condenser  is  to  be  made  out  of 
sheets  of  mica  12  x  25  cm.,  .3  mm.  thick,  and  coated  on  one 
side  with  a  very  thin  film  of  silver.  How  many  sheets  are 
required  ?  The  relative  permittivity  of  the  mica  is  equal  to  about 
6.  Ans.  About  96  sheets,  48  sheets  in  parallel  per  terminal. 

PROB.  119.  Let  the  dielectric  in  problem  108  consist,  instead 
of  glass,  of  three  layers  of  different  materials.  Let  the  thick- 
nesses of  these  layers  be  1.2,  .7,  and  i.i  mm.,  and  let  the  corre- 
sponding values  of  relative  permittivities  be  2,  3,  and  5.  What 
is  the  capacity  of  the  condenser  ?  Solution  :  The  layers  being 
divided  by  equipotential  surfaces,  infinitely  thin,  insulated  metal 
plates  may  be  imagined  between  the  separate  materials,  without 
changing  in  any  way  the  stresses  or  the  displacements.  The 
problem  is  thus  reduced  to  that  of  calculating  the  equivalent 
permittance  of  three  permittances  in  series.  The  component 
permittances,  per  unit  area  of  the  plates,  are  :  2Ka/.i2  ;  3*^.07  ; 
5Ka/.n.  Hence,  according  to  eqs.  (70)  and  (71),  we  have  that 
C=  Ka  x  50  x  7o/(.i2/2  +  .07/3  +  .11/5)  =  -00294  mf. 

PROB.  120.  The  condenser  described  in  the  preceding  problem 
is  subjected  to  a  difference  of  potential  equal  to  10  kv.  What 
are  the  voltage  gradients  (stresses)  in  the  separate  dielectrics  ? 


or  be  displaced  ;  their  reciprocals,  resistance  and  elastance,  show  the  degree 
of  difficulty  to  the  passage  of  electricity.  Similarly,  the  coefficient  I|K 
should  be  called  the  elastivity  of  the  medium,  being  analogous  to  resistivity 
p,  and  to  the  modulus  of  elasticity  in  mechanics.  Elastance  of  dielectrics 
could  be  properly  measured  in  units  called  daraf s,  one  daraf  being  the 
reciprocal  of  one  farad  ;  it  is  therefore  designated  by  the  same  name  spelled 
backwards  (same  as  ohm  and  mho).  As  the  farad  is  too  large  a  unit,  so 
the  daraf  is  too  small  a  unit  for  practical  purposes  ;  elastances  are  conven- 
iently measured  in  megadarafs,  same  as  insulation  resistance  is  measured  in 
megohms.  One  megadaraf  is  the  reciprocal  of  one  microfarad.  There  is 
little  doubt  that  with  the  extended  use  of  extra-high  voltages  there  will  be 
more  and  more  need  for  electrostatic  calculations,  and  for  a  scientific  design 
of  insulation  ;  then  some  such  names  and  units  will  have  to  be  agreed  upon. 


CHAP.  6.]  THE  ELECTROSTATIC  CIRCUIT.  73 

Solution  :  Voltage  drops  across  condensers  in  series  are  to  each 
other  as  their  elastances  (note  analogy  to  the  case  of  resistances 
in  series).  Hence,  denoting  the  drops  by  x^  x2J  x31  we  have 
xl  :  x2  :  xt  =  .12/2  :  .07/3  :  .11/5.  Besides,  xl  -f  x,  -f  x3  =  10. 
Solving  these  equations  we  find  xl  =  s.ykv.;  x2  =  2.21  kv. ;  xs  = 
2.09  kv.  The  corresponding  potential  gradients  are  4.75,  3.16, 
and  1.9  kv.  per  mm.  of  thickness. 

PROB.  121.  Deduce  a  general  expression  for  the  capacity  of  a 
single-core,  lead-covered  cable  (Fig.  14),  L  miles  long.  The 
capacity  is  understood  here  as  the  permittance  between  the  con- 
ductor and  the  sheathing.  Solution  :  The  elastance  (reciprocal 
of  permittance)  of  an  infinitesimal  concentric  layer  of  thickness 
dx  and  of  the  length  /  inches  is  dx\{Wfxl)t  so  that  the  elastance 
of  the  whole  dielectric  is 


f 

J  ai 


=  (I/27TK/)  Ln  ajar     .     .      (72) 


The  permittance  of  the  cable  is  the  reciprocal  of  its  elastance. 
Using  common  logarithms,  we  get 

C=  27TK//[2.3026  log  aja^.     .     .     ...    (73) 

In  the  English  system  K  =  .2244  K  X  lo"6  where  A^is  the  rela- 
tive permittivity  of  the  insulation  used.  The  length  in  inches 
/=  12  x  528oZ,.  Substituting,  we  get 

C=  .o388A"Z,/log  K/O,  (in  mf.).     .     .     .      (74) 

PROB.  122.  What  is  the  capacity  per  mile  of  the  cable  given 
in  problem  1 1 1 ,  and  what  is  the  specific  inductive  capacity  of 
the  dielectric?  Ans.  C  =  .1583  mf./mile  ;  K—  2.73. 

PROB.  123.  What  is  the  maximum  and  the  average  stress  in  the 
cable  in  problem  in  (use  answers  to  problems  in  and  122)? 
Solution  :  Fmax  =  DmJK '=  .0467/02244  x  lo"6  X  2.73)  =  76.2 
kv./inch.  Fm=*  1 2^375  =  32  kv./inch.  NOTE  :  The  results  derived 
in  this  problem  are  of  great  practical  importance.  They  cor- 
roborate the  statement  made  above  that  unless  the  thickness  of 
a  dielectric  is  small  as  compared  to  the  radius  of  curvature,  the 
average  stress  gives  no  idea  of  the  maximum  flux  density  at  the 
most  dangerous  point. 


74  THE  ELECTROSTATIC  CIRCUIT.  [CHAP.  6. 

PROB.  124.  What  is  the  general  expression  for  the  stress  in 
the  insulation,  in  the  cable  shown  in  Fig.  14,  at  a  distance  x 
from  the  center,  when  the  voltage  between  the  core  and  the 
sheathing  is  equal  to  E^  Solution  :  The  dielectric  flux  density, 
and  consequently  the  stress,  is  inversely  proportional  to  the 
distance  from  the  center  of  the  cable.  The  stresses  are  also 
proportional  to  the  applied  voltage.  Hence,  F  —  NE\x,  where 
Nis  a  constant.  This  constant  is  determined  from  the  condition 
that  the  applied  voltage  is  the  line  integral  of  the  stress,  see  eq. 
(64).  We  have 

E  =  C"~Fdx  =  NE  C**dxlx  =  NE  I^n  («,/«,). 

•^    a\  *J  a\ 

Or,  N=  (Ln  ajaj-1,  so  that  F  =  EKxI&aJaJ. 

PROB.  125.  Check  the  solution  of  problem  123  using  the  result 
of  the  preceding  problem. 

PROB.  126.  Show  by  actual  calculation  that  in  the  foregoing 
cable,  the  maximum  stress  in  the  dielectric  is  reduced  by  using 
conductor  No.  i  B.  &  S.  instead  of  No.  4,  with  the  same  diameter 
of  the  sheathing.  This  result  is  obtained  in  spite  of  the  fact  that 
the  insulation  becomes  thinner,  and  consequently  has  a  larger 
average  stress. 

PROB.  127.  Referring  to  the  preceding  problem,  show  that  it 
is  of  advantage  to  make  the  ratio  of  ajav  about  equal  to  c,  where 
c  =  2.71828  ....  is  the  base  of  natural  logarithms.  When  the 
diameter  of  the  conductor  is  further  increased,  so  that  the  ratio 
of  ajdi  becomes  less  than  «,  a  further  increase  in  the  diameter  of 
the  core  increases  the  maximum  stress,  instead  of  reducing  it. 
Solution  :  The  stress  at  the  core  is 


according   to   problem    124.     F1   reaches  its  maximum  with  a 
variable  a^  when  dFJdal  =  o.     Differentiating,  we  get 

dFJda,  =  E  [i  -  Ln  («>,)]/  [«,  I,n(«,/«i)]'  =  o  ; 

whence,   i  —  Ln  (ajaj  =  o,  or  ajav  =  c. 

PROB.  128.  The  insulation  of  a  cable  consists  of  two  layers, 
the  outside  radius  of  the  inner  layer  being  b.  The  relative  per- 
mittivities of  the  materials  are  K^  and  Ky  Extend  the  expression 


CHAP.  6.]  THE  ELECTROSTATIC  CIRCUIT.  75 

(74)  for  the  capacity  to  this  case.  Solution  :  The  two  permit- 
tances 

£;  =  .0388  A^/log  (#O 
and 

C2  =  .0388  A-2Z,/log  (ajb} 

are  in  series.  Therefore,  the  resultant  permittance,  according 
to  formula  (71),  is 

c=  .03881, 

The  result  can  evidently  be  extended  to  any  number  of  concentric 
layers  of  insulation. 

Note  to  page  67.  The  author  considers  the  above- given  value 
of  K  for  air  to  be  an  experimental  coefficient,  in  the  same  sense 
in  which  other  properties  of  materials  are  characterized  by 
experimental  coefficients.  This  is  because  for  an  engineer  the 
volt  and  the  ampere  are  arbitrary  units  established  by  an  inter- 
national agreement,  no  matter  what  their  relation  to  the  so- 
called  absolute  units. 

The  value  of  K  can  be  calculated  theoretically,  assuming  the 
ratio  between  the  electrostatic  and  the  electromagnetic  units  to 
be  known.  Namely,  in  the  absolute  electrostatic  system  of  units 
a  plate  condenser  having  an  area  of  A  sq.  cm.  and  a  distance 
between  the  plates  equal  to  /  cm.,  has  a  capacity  equal  to  A\^irl. 
The  factor  4^  enters  on  account  of  an  unfortunate  selection  of 
the  expression  for  Coulomb's  law  ;  namely  it  should  have 
been  ee'l^rr"1,  instead  of  ee'lr*.  In  the  absolute  electromagnetic 
units  the  same  capacity  is  equal  to  (A  fair!)  (3  x  iolo)~2,  where 
3  x  io10  is  the  velocity  of  light  in  centimeters  per  second.  To 
obtain  the  result  in  microfarads,  the  foregoing  expression  must 
be  multiplied  by  io15.  On  the  other  hand,  the  same  capacity, 
expressed  in  the  rational  units,  is  =  K.AII.  Equating  the  two 
expressions,  gives  K  =  io~~5/(9  X  47r)  =  .08842  x  io~6  microfarads 
(microcoulombs  per  volt)  per  square  centimeter  cross- section,  per 
centimeter  length. 

The  fact  that  K  can  be  expressed  through  the  velocity  of  light 
does  not  make  K  the  less  an  empirical  coefficient,  because  the 
velocity  of  light  itself  is  determined  experimentally.  As  a  matter 
of  fact,  one  of  the  ways  in  which  the  velocity  of  light  is  deter- 
mined consists  in  calculating  it  indirectly  from  the  value  of  K 
obtained  from  measurements. 


CHAPTER  VII. 

THE  ELECTROSTATIC  CIRCUIT. 
{Continued). 

24.  Capacity  in  Alternating-Current  Circuits.  The  effect 
of  electrostatic  capacity  in  alternating-current  circuits  is  dis- 
cussed in  detail  in  Arts.  436  to  445  of  the  author's  Experimental 
Electrical  Engineering  (Arts.  127  to  136  in  the  first  edition.) 
It  is  shown  there  that  an  electrostatic  capacity  or  permittance 
gives  rise  to  a  charging  current,  which  leads  the  voltage  at  the 
terminals  of  the  condenser  by  90  degrees  in  phase.  The  value 
of  the  charging  current  is 

7=^.27r/.Cx  io~6  .....      (75) 

where/"  is  the  frequency  in  periods  per  second,  and  C  is  the 
capacity  in  microfarads.  The  product  2irfCx  lo"6  has  the  di- 
mension of  a  mho,  because  it  is  equal  to  the  ratio  of  a  current 
to  a  voltage.  By  analogy  with  the  inductive  susceptance 
introduced  in  Art.  15,  the  product  2-rr/Cx  io~6  is  called  the 
capacity  susceptance,  and  is  also  measured  in  mhos.  Some  authors 
call  it  capacitance. 

The  inductive  susceptance  causes  the  current  to  lag  90°  be- 
hind the  applied  voltage,  while  the  capacity  susceptance  makes 
it  lead  by  90°.  Therefore,  in  the  analytical  treatment,  given  in 
Chapter  V,  it  is  necessary  to  consider  the  susceptance  negative, 
when  it  is  caused  by  a  condenser,  and  positive  when  it  is  caused 
by  a  reactive  coil.  We  have,  therefore, 

£=_27r/Cx  lo-6.    .-;-..,'    .-  .     .      (76) 

The  reciprocal  of  a  susceptance  is  a  reactance,  and  we  have,  as 
in  Art.  n  that  the  capacity  reactance 

*  =  -  io6/27i/C. (77) 

It  must  be  remembered  that  a  reactance  is  numerically  equal  to 
the  reciprocal'  of  a  susceptance  only  when  no  ohtnic  resistance  is 
present.  Otherwise  eqs.  (45)  and  (46)  must  be  used.  The  use 
of  the  capacity  susceptance  and  of  the  capacity  reactance  is 
made  clearer  in  the  solution  of  the  examples  that  follow. 


CHAP.  7-]  THE  ELECTROSTATIC  CIRCUIT.  77 

PROB.  129.  A  condenser  of  7.3  mf.  permittance  is  connected 
across  a  500  v.,  60  cycle  supply.  What  is  the  susceptance  and 
the  charging  current?  Ans.  b  =  —  .002754  mho  ;  7=71.377 
amp.  the  voltage  being  the  reference  vector. 

PROB.  130.  The  condenser  in  the  preceding  problem  is  shunted 
by  a  non-inductive  resistance  of  750  ohm.  Find  the  total  current 
and  the  power  factor.  Solution  :  The  current  through  the  re- 
sistance is  =  500/750=  .6667  amp.;  tan  <£=  r.377/,6667  =  2.065; 
cos<£  =  43.58  per  cent  (leading).  Total  current  =  .6667/.435S 
=  1.53  amp. 

PROB.  131.  The  condenser  and  the  resistance  in  the  preceding 
problem  are  connected  in  series,  instead  of  in  parallel.  What  is 
the  equivalent  parallel  combination?  Ans.  Cp  =  1.387  mf.; 
rp  —  926  ohm. 

PROB.  132.  A  magnetic  reactance  of  65  ohm  is  connected  in 
parallel  with  a  capacity  of  73.6  mf.,  across  a  2200  v.,  25  cycle 
circuit.  Determine  the  total  current,  and  the  component 
currents,  through  the  reactance  and  through  the  condenser. 
Ans.  33.85  —  25.44  =  8-41  amp.  (lagging).  This  is  a  case  of 
partial  current  resonance,  the  total  current  being  smaller  than 
one  of  its  components. 

PROB.  133.  The  condenser  and  the  reactance  coil  given  in  the 
preceding  problem  are  connected  across  the  same  line  in  series, 
instead  of  in  parallel.  Find  the  total  current  and  the  component 
voltages.  Ans.  102.3  amp.  (leading);  2200  =  8850  —  6650  volts. 
This  is  a  case  of  partial  voltage  resonance,  the  voltage  drop 
across  one  of  the  two  devices  being  larger  than  the  applied 
voltage. 

PROB.  134.  The  voltage  at  the  receiver  end  of  a  25  cycle  single- 
phase  transmission  line  is  45  +757  kv. ;  the  load  current  is 
178  +769  amp.  The  impedance  of  the  line  is  32  -f/68  ohm  ; 
the  capacity  of  the  line  is  4.24  mf.  Calculate  the  generator  cur- 
rent and  voltage.  For  the  purposes  of  calculation  one  half  of 
this  capacity  can  be  assumed  to  be  connected  across  the  generator 
end  of  the  line,  the  other  half  across  the  receiver  end.  Solution  : 
The  capacity  susceptance  at  the  receiver  end  of  the  line  in  mhos  is 
—  27rX  25X2.I2X  io~6=—  .333  X  io~~3.  The  corresponding  charg- 
ing current  is/333  x  io~3  (45000  +757000)  =  —  19  +715  amp. 


78  THE  ELECTROSTATIC  CIRCUIT.  [CHAP.  7. 

Consequently,  the  total  line  current  is  159  +784  amp.  The  line 
drop  is  (159  +784)  (32  +768)  =  —  624  +713500  volt.  Generator 
voltage  =  44.38  +770.5  kv.  Charging  current  at  the  generator 
end  =  7.333  (44-38  +770.5)  =  —  23.5  +714.79  amp.  Generator 
current  =  135.5  +798.8  amp. 

PROB.  135.  Indicate  the  general  method  of  solution  of  the  pre- 
ceding problem  in  the  abbreviated  symbolic  notation.  Solution  : 
Let  the  capacity  susceptance  at  each  end  of  the  line  be  t>,  where 
b  is  a  negative  quantity.  Let  the  receiver  or  the  load  voltage 
be  E^  .  Then,  the  charging  current  at  the  receiver  end  is  7'c  = 

—jbE^  ,  and  the  line  current  is  =  7L  +  7'c  •  If  the  line  imped- 
ance is  Z,  the  generator  voltage  EG  —  7fL  +  Z  (7L  +  /'c  ).  The 
charging  current  at  the  generator  end  is  7"c  =  —jbEG  .  Gene- 
rator current  7G  =  7L  +  7'c  +  7"c  . 

25.  Energy  in  the  Electrostatic  Field.  When  a  condenser 
is  being  charged  a  current  flows  into  it  from  the  source  of  elec- 
tromotive force.  This  involves  the  expenditure  of  a  certain 
amount  of  energy,  necessary  to  produce  the  required  stresses 
and  strains  in  the  dielectric  of  the  condenser.  This  energy  is 
not  converted  into  heat,  and  therefore  lost,  as  in  the  case  of 
metallic  conduction  :  The  energy  is  stored  in  the  potential  form 
in  the  dielectric,  and  can  be  returned  to  the  circuit  by  reducing 
the  voltage  at  the  condenser  terminals.  With  reference  to  the 
analogy  shown  in  Fig.  12,  the  energy  expended  by  the  pump  in 
straining  the  partition  is  stored  in  the  partition  by  virtue  of  its 
elastic  character,  and  exists  coincident  with  the  stresses  and 
strains.  It  can  be  returned  to  the  piston  rod  by  allowing  the 
piston  to  be  moved  by  the  elastic  partition. 

It  is  necessary  in  some  cases  to  calculate  the  energy  stored  in 
an  electrostatic  field  ;  or  at  least  to  represent  the  energy  stored 
per  cubic  inch  or  centimeter  of  dielectric  as  a  function  of  the 
stress  F,  strain  D,  and  the  permittivity  K,  at  the  point  under 
consideration. 

Consider  first  the  simplest  case  of  a  plate  condenser  (Fig.  10), 
and  neglect  the  small  amount  of  displacement  occurring  outside 
the  space  between  the  plates.  Let  the  condenser  be  charged  by 
gradually  raising  the  voltage  at  its  terminals  from  o  to  a  final 


CHAP.  7.]  THE  ELECTROSTATIC  CIRCUIT.  79 

value  E ;  let  e  and  i  be  the  instantaneous  values  of  the  voltage 
and  of  the  charging  current  at  a  moment  /  during  the  process  of 
charging.*  The  total  electrical  energy  delivered  to  the  con- 
denser in  charging  is 


=      f  Cidt-    £Te.dq (78) 

•so  «/   o 


T 

W 

o 


where  T  is  the  total  time  of  charging,  and  dq  =  idt  is  the  infini- 
tesimal charge  or  displacement  added  to  the  condenser  during 
the  interval  of  time  dt.  The  quantities  dq  and  e  can  be  expressed 
through  the  instantaneous  flux  density  Z?t  and  the  stress  JFt  from 
eqs.  (59)  and  (60).  Performing  the  substitution,  and  taking 
the  constant  quantities  A  and  /  outside  the  sign  of  integration, 
we  get 

T  .     .     .     .     .     .      (79) 


In  order  to  integrate  this  expression  Z?t,  must  be  expressed 
through  jpt,  or  vice  versa.  The  relation  between  the  two  is 
given  by  eq.  (65).  Eliminating  Z?t  we  obtain  : 

W=  *Al  C  FtdFt  =  %KvF*  (80) 

V     O 

where  v  —  Al  is  the  volume  of  the  dielectric,  and  F  is  the  final 
value  of  the  stress,  at  the  time  T.  Hence,  the  energy  stored  per 
unit  volume  of  dielectric  is 

W\v  =  \*F\     .     .     .     ..    :     „      (81) 

The  ratio  W\v  is  called  the  density  of  energy.    Using  the  relation 

D  =  K  F,  eq.  (81  )  can  also  be  written  in  the  following  two  forms  : 

W\v  =  \FD  =  \D*\K.     .     ....    r    (82) 

The  analogy  with  formula  (13)  in  Art.  4  is  apparent  at  once. 

The  stored  energy  can  also  be  expressed  through  the  permit- 
tance or  capacity  of  the  dielectric.  We  have  from  eq.  (68) 
dq  —  C.de  ;  substituting  into  (78)  and  integrating  we  get 

(83) 


*The  voltage  aud  the  charging  current  rise  gradually,  even  though  the 
key  K'\s  closed  suddenly.  This  is  on  account  of  an  ever-present  inductance 
in  the  leads  which  inductance  acts  as  a  kind  of  electromagnetic  inertia. 


8o  THE  ELECTROSTATIC  CIRCUIT,  [CHAP.  7. 

Since  the  final  charge,  or  total  displacement,  Q  =  CE,  the  energy 
can  be  also  represented  in  the  following  two  forms  : 

(84) 


These  expressions  are  analogous  to  eq.  (ya)  in  Art.  2. 

Let  now  the  condenser  be  of  an  irregular  form,  as  shown  in 
Fig.  13.  The  stress  F  and  the  displacement  D  are  different  at 
different  points,  so  that  it  is  necessary  to  consider  infinitesimal 
layers  of  the  dielectric  between  consecutive  equipotential  sur- 
faces, and  infinitesimal  threads  of  displacement  between  the 
electrodes.  Consider  an  infinitesimal  volume  mnpq  of  the  die- 
lectric, comprised  of  a  filament  HH1  ',  between  two  equipo- 
tential surfaces  MN  and  M'N'.  The  sides  mp  and  nq  can 
be  provided  with  infinitely  thin  metal  films,  because  these  sides 
lie  in  the  equipotential  surfaces  so  that  no  current  could  flow 
along  these  metal  coatings.  Then  the  element  of  volume  under 
consideration  is  converted  into  a  small  plate  condenser  ;  the  flux 
density  and  the  stress  within  this  element  can  be  considered 
uniform,  so  that  formula  (80)  holds  true,  and  we  have 

\*F'1-dv  .....      .  •    (85) 


Differentials  are  used  because  both  the  volume  and  the  energy 
stored  are  infinitesimal.     The  density  of  energy 


....     (86) 

has  the  same  expression  as  in  the  case  of  a  plate  condenser  ;  but 
its  numerical  value  is  different  from  point  to  point,  since  F  is 
variable.  The  two  other  expressions  for  the  density  of  energy 

.     .     '.'     ,      (87) 


also  ho^d  true  for  the  points  of  a  non-uniformly  stressed  dielectric, 
provided  that  proper  values  of  D  and  .Fare  used  for  each  point. 
The  total  energy  stored  in  a  non-uniform  electrostatic  field  is 
equal  to 

.    .    (88) 

Here  F  and  D  must  be  given  as  functions  of  co-ordinates,  and 
the  integration  extended  over  the  whole  space  occupied  by  the 
field.  Eqs.  (83)  and  (84)  are  true  for  condensers  of  any  shape, 


CHAP.  7.]  THE  ELECTROSTATIC  CIRCUIT.  8 1 

because  in  the  deduction  of  these  formulae  no  assumption  was 
made  as  to  the  particular  form  of  the  dielectric,  or  of  the 
electrodes. 

The  expressions  for  the  electrostatic  energy  of  the  field, 
derived  above,  are  identical  with  the  expressions  for  the  potential 
energy  of  stressed  elastic  bodies,  and  this  is  consistent  with  the 
assumed  structure  of  the  dielectric,  shown  in  Fig.  u.  Namely, 
consider  the  work  necessary  to  strain  the  elastic  fibres  per  one 
cubic  centimeter  of  the  material.  During  the  process  of  shearing 
the  stress  varies  from  zero  to  its  final  value  F.  Let  Ft  be  some 
intermediate  value,  and  Dt  the  corresponding  strain.  While  the 
strain  increases  from  Z7t  to  (Di  -f-  dD^  the  stress  Ft  may  be  con- 
sidered constant  ;  the  infinitesimal  work  of  shearing  is  therefore 
equal  to  F^dDt.  The  total  work  of  shearing 


But,  according  to  Hooke's  law  of  elasticity,  shearing  strains  are 
proportional  to  stresses,  so  that  a  relation  exists  between  D  and 
F,  similar  to  eq.  (65).  We  thus  arrive  again  at  the  result  that 
the  work  necessary  to  strain  one  cubic  unit  of  an  elastic  material 
is  equal  to  \  K  F3. 

PROB.  136.  Calculate  the  total  stored  energy,  and  the  density  of 
energy  in  the  condenser  given  in  problem  108.  Ans.  20.52 
milliwatt-seconds  (millijoules);  19.53  microjoules  per  cubic 
centimeter. 

PROB.  137.  Assuming  the  relative  permittivity  of  the  insula- 
tion given  in  problem  1 10  to  be  2.5  ;  what  is  the  density  of  energy 
at  which  the  material  is  broken  down?  Ans.  \K.F2  =  .0725 
joules  per  cu.  inch. 

PROB.  138.  Show  that  in  a  single-core  cable  (Fig.  14)  the 
density  of  energy  in  the  insulation  varies  inversely  as  the  square 
of  the  distance  from  the  center  of  the  core. 

PROB.  139.  Show  that  in  a  spherical  condenser  the  density  of 
energy  varies  inversely  as  the  fourth  power  of  the  distance  from 
the  center. 

PROB.  140.  Check  in  problem  in  that  the  energy  supplied  to 
the  cable  from  the  source  is  equal  to  that  stored  in  the  dielectric. 
The  specific  inductive  capacity  of  the  insulation  is  2.73  (see 


82  THE  ELECTROSTATIC  CIRCUIT.  [CHAP.  7. 

answer  to  problem   122).     Solution:    Max.   density  of  energy, 


joules  per  cu.  inch.  The  density  of  energy  at  a  distance 
x  from  the  center  is  17.8  x  icr~4  (ajxft  since  the  energy 
density  varies  inversely  as  the  square  of  the  distance  from  the 
center.  The  energy  in  an  infinitesimal  concentric  layer  of  thick- 
ness dx,  and  one  inch  long,  is  dW=  17.8  X  io~~4  (ajx}2  •  znxdx. 
Integrating  between  the  limits  al  and  #2  gives  the  total  energy 
per  inch  length  of  the  cable,  W  =  1.798  x  io~4  joules.  The 
energy  per  mile  is  11.39  joules.  On  the  other  hand,  according 
to  eq.  (84),  the  same  energy  is  equal  to  \  X  .0019  X  12000=11.4 
joules. 

26.  The  Electrostatic  Corona.  When  the  voltage  is  raised 
sufficiently  high  at  the  terminals  of  an  air-condenser,  a  pale  violet 
light  appears  at  the  edges,  at  the  sharp  points,  and  in  general  at 
the  protruding  parts  having  a  comparatively  small  radius  of 
curvature.  This  silent  discharge  into  air,  due  to  an  excessive 
electrostatic  flux  density,  is  called  the  electrostatic  corona.  In  the 
regions,  where  the  corona  appears,  the  air  is  electrically  '  '  broken 
down  '  '  and  ionized  so  that  it  becomes  a  conductor  of  electricity. 
When  the  voltage  is  raised  still  higher  the  so-called  brush  dis- 
charge takes  place,  until  the  whole  thickness  of  the  dielectric  is 
broken  dow-n,  and  a  disruptive  discharge,  or  spark,  jumps  from 
one  electrode  to  the  other. 

When  the  electrodes  have  projecting  parts  or  sharp  edges  the 
corona  is  formed  at  a  voltage  far  below  that  at  which  the  dis- 
ruptive discharge  occurs  ;  the  operating  voltage  of  such  devices 
is  generally  limited  to  that  at  which  the  corona  forms.  No 
corona  is  usually  permissible  in  regular  operation,  first,  because 
it  involves  a  considerable  loss  of  power  ;  second,  because  the  dis- 
charge, if  allowed  to  play  on  some  other  insulation,  will  soon 
char  and  destroy  it.  There  are  cases,  however,  in  which  some 
corona  formation  is  harmless  :  when  the  air  which  is  broken 
down  becomes  a  part  of  the  electrode,  smooths  down  the  shape 
of  the  protruding  parts,  increases  their  area,  and  thus  reduces 
the  dangerous  flux  density,  making  it  more  uniform. 

The  formation  of  corona  must  be  kept  in  mind  in  the  design 
of  high-tension  insulation,  and  in  high-potential  tests.  Shapes 
and  combinations  of  parts  should  be  avoided  which  lead  to  high 


CHAP.  7.]  THE  ELECTROSTATIC  CIRCUIT.  83 

or  non-uniform  dielectric  flux  densities.  Fig.  13  shows  the 
reason  why  the  dielectric  flux  density  or  the  potential  gradient 
is  higher  near  protruding  parts.  The  equipotential  surfaces,  for 
obvious  geometrical  reasons,  lie  closer  to  each  other  near  such 
parts,  because  at  a  reasonable  distance  from  the  electrodes  the 
shape  of  the  equipotential  surfaces  is  not  affected  by  small 
irregularities  in  the  shape  of  the  metallic  parts. 

Corona  is  formed,  in  other  words  air  is  broken  down,  at  a 
dielectric  flux  density  Z?max  of  about  .00238  microcoulombs  per 
square  centimeter,  or  .0153  me.  per  sq.  inch  (A.  C.,  effective 
value,  sine- wave  voltage).  This  flux  density  corresponds  to  a 
potential  gradient  or  stress  faM  =  Dmjn  =  26.8  effective  kilovolts 
per  centimeter  (68.1  kv.  per  inch).  In  practice,  it  is  advisable 
to  keep  somewhat  below  these  figures. 

PROS.  141.  A  conductor  i  inch  in  diameter  is  surrounded  by 
a  concentric  metal  cylinder,  6  inch,  inside  diameter.  What 
alternating  voltage  can  be  allowed  between  the  cylinder  and  the 
conductor  at  a  factor  of  safety  of  1.5  against  the  formation  of 
the  corona  ?  Solution  :  The  permissible  flux  density  at  the 
surface  of  the  conductor  is  .0153/1.5  =  .01023  me.  per  sq.  inch. 
Hence,  the  charge  per  inch  of  axial  length  must  not  exceed 
TT  x  i  X  .01023  =  .0322  me.  According  to  formula  (68),  the 
voltage  is  equal  to  the  charge  divided  by  the  permittance  of  the 
dielectric.  The  permittance  is  calculated  in  this  case  according 
to  eq.  (73)  and  is  equal  to  .7865  X  lo""6  m.f.  per  inch  length. 
Consequently,  the  permissible  voltage 

E=      .0322/07865  X   lo"6)  lio-3  =  40.8  kv. 

PROB.  142.  Show  that  in  the  preceding  problem  the  maximum 
permissible  voltage  is  proportional  to  the  expression  «,  log  (aja^), 
see  Fig.  14.  Plot  a  curve  giving  values  of  maximum  permissible 
voltage,  against  diameters  of  the  inner  conductor  as  abscissae. 
The  diameter  of  the  outside  cylinder  is  assumed  constant  and 
equal  to  6  inches.  Ans.  The  curve  reaches  a  maximum  =  50.3 
kv.  when  the  diameter  of  the  conductor  is  2.21  inch. 

PROB.  143.  The  charging  current  per  mile  of  wire  in  a  trans- 
mission line  is  .  i  amp.  Given  that  the  conductors  consist  of  No. 
.0000  wire  B.  &  S.  and  that  the  frequency  is  25  cycles  per  second, 
determine  the  factor  of  safety  of  the  line  against  the  corona. 


84  THE  ELECTROSTATIC  CIRCUIT.  [CHAP.  7. 

Solution :  The  line  is  charged  during  one  quarter  of  a  cycle, 
or  during  i/ioo  of  a  second.  The  average  current  is  (.1  x  ^2) 
X  (2/7r)  =  .09  amp.  Maximum  charge  is  .09  x  (i/ioo)  x  io6 
=  900  me.  per  mile.  The  area  of  the  wire  per  mile  is  91600  sq. 
inch.,  so  that  the  flux  density  is  900/91600  =  .00983  me.  per 
sq.  inch.  The  factor  of  safety  is  .oi53/. 00983  =  1.56. 

27.  Dielectric  Hysteresis  and  Conduction.  When  an  alter- 
nating voltage  is  applied  at  the  terminals  of  a  condenser,  the 
dielectric  is  subjected  to  periodic  stresses  and  strains.  If  the 
material  were  perfectly  elastic,  no  energy  would  be  lost  during 
one  complete  cycle,  because  the  energy  stored  during  the  periods 
of  increase  in  voltage  would  be  given  up  to  the  circuit  when  the 
voltage  decreased.  In  reality,  there  are  reasons  to  believe  that 
the  electric  elasticity  of  solid  and  liquid  dielectrics  is  not  perfect. 
Referring  to  Fig.  n,  dielectrics  behave  as  if  there  was  some 
kind  of  friction  between  the  filaments,  so  that  the  applied  volt- 
age has  to  overcome  this  friction,  in  addition  to  the  elastic 
forces.  The  work  done  against  friction  is  converted  into  heat, 
and  is  lost,  as  far  as  the  circuit  is  concerned.  The  phenomenon 
is  similar  to  the  familiar  magnetic  hysteresis,  and  is  therefore 
called  dielectric  hysteresis.  The  energy  lost  per  cycle  is  pro- 
portional to  the  square  of  the  applied  voltage,  because  both  the 
displacement  and  the  stress  are  proportional  to  the  voltage. 

Under  normal  conditions,  that  is  to  say  when  stresses  are  well 
below  the  ultimate  stress,  the  loss  of  power  in  dielectric  hysteresis 
is  exceedingly  small.  Some  investigators  even  doubt  its  exist- 
ence at  all.  There  is  often  an  appreciable  loss  of  power  in  com- 
mercial condensers,  but  this  loss  can  be  mostly  attributed  to  the 
fact  that  dielectrics  are  not  perfect  insulators.  While  their 
ohmic  resistance  is  exceedingly  high,  as  compared  to  that  of 
metals,  they  nevertheless  conduct  some  current,  especially  at 
high  voltages.  Thus,  the  observed  loss  of  power,  and  the  heat- 
ing of  condensers,  can  be  simply  ascribed  to  the  PR  loss  in  the 
insulation.  Moreover,  small  amounts  of  corona  can  form  at  the 
edges  and  projecting  parts,  even  at  the  operating  voltage,  and 
thus  be  an  additional  source  of  loss.  Some  small  loss  is  also  due 
to  the  ohmic  resistance  of  the  metallic  sheets  which  compose  the 
condenser. 


CHAP.  7.]  THE  ELECTROSTA  TIC, .  CflRCU'lT.  "85 

An  imperfect  condenser,  that  is  to  say  a  condenser  which 
shows  a  loss  of  power,  from  one  cause  or  another,  can  be  replaced, 
for  purposes  of  calculation,  by  a  perfect  condenser  with  an 
ohmic  resistance  shunted  around  it.  The  resistance  is  selected 
of  such  a  value  that  the  PR  loss  in  it  is  equal  to  the  loss  of 
power  from  all  the  causes  in  the  given  imperfect  condenser. 
The  actual  current  through  the  imperfect  condenser  is  consid- 
ered then  as  consisting  of  two  components :  The  wattless  com- 
ponent, flowing  through  the  ideal  condenser,  and  the  loss  com- 
ponent, in  phase  with  the  voltage,  flowing  through  the  shunted 
resistance.  In  this  way,  imperfect  condensers  can  be  treated 
graphically  or  analytically,  according  to  the  ordinary  laws  of  the 
electric  circuit. 

PROB.  144.  A  certain  kind  of  condensers  shows  a  loss  of  power 
of  about  17.9  watts  per  microfarad,  at  about  2200  v.,  25  cycles. 
By  what  fictitious  resistance  should  an  ideal  condenser  be  shunted 
in  order  to  replace  a  condenser  of  this  kind,  having  a  capacity 
of  1.5  mf.?  Ans.  About  .18  megohm. 


LITERATURE. 

ALTERNATING  CURRENT  PHENOMENA,  by  Chas.  P.  Steinmetz. 
THEORETICAL  ELEMENTS  OF  ELECTRICAL  ENGINEERING,  by 
Chas.  P.  Steinmetz. 

ALTERNATING  CURRENTS,  by  Bedell  and  Crehore. 
VECTORS  AND  VECTOR  DIAGRAMS,  by  Cramp  and  Smith. 
ELECTRICAL  ENGINEERING,  by  Thomalen. 

DIE   WlSSENSCHAFTLICHEN    GRUNDLAGEN  DER  ELEKTROTECH- 

NIK,  by  Benischke. 

DIE  WECHSELSTROMTECHNIK,  Vol.  I,  by  Arnold  and  la  Cour. 
PROBLEMS  IN  ELECTRICAL  ENGINEERING,  by  Waldo  V.  L,yon. 
ELECTRICAL  PROBLEMS,  by  Hooper  and  Wells. 
THE  ELEMENTS  OF  ELECTRICAL  ENGINEERING,  Vol.   II,  by 

Franklin  and  Esty. 

MODERN  VIEWS  OF  ELECTRICITY,  by  Oliver  Lodge. 
ELECTRIC  WAVES,  by  W.  S.  Franklin. 

ELEMENTS  OF  ELECTROMAGNETIC  THEORY,  by  S.  J.  Barnett. 
KAPAZITAT  UND  INDUKTIVITX.T,  by  Ernst  Orlich. 
TEXT-BOOK  OF  ELECTRICAL  MACHINERY,  by  Ryan,  Norris,  and 

Hoxie. 


AC? 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 


Return  to  desk  from  which  borrowed. 

*' 


This  book  is  DUE  on  the  last  date  stamped  below. 

£NG  NEERING  LIBRARY 


APR  T  0  1950 


LD  21-100m-9,'48(B399sl6>476 


